This page is ideal for practising all the skills of differentiation. You may wish to use this page in preparation for a test on this topic or for the final examinations. The quizzes on this page have been carefully created to take you through all the skills that you need. If you want a more in depth look, then you should go to the individual pages on these topics.
On this page, you can revise
- differentiating xn
- differentiating f(x) + g(x)
- differentiating sinx, cosx, ex and lnx
- the chain rule
- the product rule
- the quotient rule
- related rates of change
- differentiating using a graphical display calculator
- graphs and the gradient function
- stationary points
- equations of tangents and normals
- kinematics
Here's a quiz that practises differentiating xn
START QUIZ!
Differentiating x to the n (2nd) 1/1
If \(\large\frac{\mathrm{d}}{\mathrm{d}x}(2x^3)=ax^b\), find a and b
a =
b =
Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(2x^3)=6x^2\)
If \(\large y = 3x^5\), then \(\large\frac{\mathrm{d}y}{\mathrm{d}x}=ax^b\)
Find a and b
a =
b =
Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)
If \(f(x)=2x^{-1}\), then \(f'(x)=ax^{b}\)
a =
b =
Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)
Find \(\large\frac{\mathrm{d}}{\mathrm{d}x}(5x)\)
What is the gradient of the straight line y = 5x ?
If \(\large A=4\pi r^2\), find \(\large\frac{\mathrm{d}A}{\mathrm{d}r}\)
\(4\pi \) is a constant
If \(\large f(x) = 10\) , what is \(\large f'(x)\) ?
The derivative of a constant is zero
If \(\large V=\frac{4}{3}\pi r^3\) , what is \(\large\frac{\mathrm{d}V}{\mathrm{d}r}\)?
Use the power rule
\(\large \frac{4}{3}\pi\) is a constant
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^2})=ax^b\)
Work out a and b
a =
b =
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^2})=\large\frac{\mathrm{d}}{\mathrm{d}x}(x^{-2})=-2x^{-3}\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2}{x^3})=ax^b\)
Work out a and b
a =
b =
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2}{x^3})=\large\frac{\mathrm{d}}{\mathrm{d}x}(2x^{-3})=-6x^{-4}\)
What is \(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3}{\sqrt{x}})\)?
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3}{\sqrt{x}})=\frac{\mathrm{d}}{\mathrm{d}x}(3x^{-\frac{1}{2}})=-\frac{3}{2}x^{-\frac{3}{2}}=-\frac{3}{2\sqrt{x^3}}\)
Here's a quiz that practises differentiating functions in the form f(x) + g(x)
START QUIZ!
Differentiating f(x)+g(x) 1/1
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(x^3+x^2)=ax^2+2x^b\)
Work out a and b
a =
b =
Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(x^3+x^2)=3x^2+2x\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(2x^4-3x^2)=8x^a+bx\)
Work out a and b
a =
b =
Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(2x^4-3x^2)=8x^3-6x\)
Work out \(\large\frac{\mathrm{d}}{\mathrm{d}x}(\sqrt{x}-\frac{1}{x})\)
Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\sqrt{x}-\frac{1}{x})=\large\frac{\mathrm{d}}{\mathrm{d}x}(x^\frac{1}{2}-x^{-1})=\frac{1}{2}x^{-\frac{1}{2}}+x^{-2}\)
If \(\large y=x(x^2-2)\) , work out \(\large\frac{\mathrm{d}y}{\mathrm{d}x}\)
There is no need to use the product rule for differentiation. Expand the brackets first
\(\large y=x(x^2-2)=x^3-2x\)
\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2-2\)
If \(\large f(x)=(x-1)(x+2)\) , work out \(f'(x)\)
There is no need to use the product rule for differentiation. Expand the brackets first
\(\large f(x)=(x-1)(x+2)=x^2+2x-x-2=x^2+x-2\)
\(\large f'(x)=2x+1\)
Work out \(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x^3-1}{x})\)
There is no need to use the quotient rule. Simple before differentiating
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x^3-1}{x})=\large\frac{\mathrm{d}}{\mathrm{d}x}(x^2- \frac{1}{x})=\large\frac{\mathrm{d}}{\mathrm{d}x}(x^2-x^{-1})\\ =2x+x^{-2}=2x+\frac{1}{x}\\ =\frac{2x^3+1}{x^2}\)
If \(\large y = 5 - 3x^2\) , work out \(\large\frac{\mathrm{d}y}{\mathrm{d}x}\) when \(\large x = \frac{1}{2}\)
\(\large\frac{\mathrm{d}y}{\mathrm{d}x}\)=
\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-6x\)
When \(\large x = \frac{1}{2}\), \(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-6\times \frac{1}{2}=-3\)
If \(\large f(x)=x^2(2x-1)\) , find \(\large f'(-1)\)
\(\large f'(-1)\) =
Expand the brackets before differentiating
\(\large f(x)=x^2(2x-1)=2x^3-x^2\)
\(\large f'(x)=6x^2-2x\)
\(\large f'(-1)=6(-1)^2-2(-1)=6+2=8\)
If \(\large y=\frac{x-4}{x^2}\) , work out \(\large\frac{\mathrm{d}y}{\mathrm{d}x}\)when x = 1
\(\large\frac{\mathrm{d}y}{\mathrm{d}x}\) =
There is no need to use the quotient rule for differentiation. Simplify before differentiating
\(\large y=\frac{x-4}{x^2}=\frac{1}{x}-\frac{4}{x^2}=x^{-1}-4x^{-2}\)
\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-x^{-2}+8x^{-3}\)
\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{x^2}+\frac{8}{x^3}\\ \)
When x = 1 , \(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{1}+\frac{8}{1}=7\)
If \(\large f(x)=2x^2-\sqrt{x}\) , find \(\large f'(0.25)\)
\(\large f'(0.25)\) =
\(\large f(x)=2x^2-x^{\frac{1}{2}}\)
\(\large f'(x)=4x-\frac{1}{2}x^{-\frac{1}{2}}=4x-\frac{1}{2\sqrt{x}}\)
\(\large f'(0.25)=4(0.25)-\frac{1}{2\sqrt{0.25}}=1-\frac{1}{2\times 0.5}=1-1=0\)
Here's a quiz that practises differentiating functions in the form sinx, cosx, ex and lnx
START QUIZ!
Differentiating sinx, cosx, ex and lnx 1/1
If \(\large y = \sin x - \cos x\) , find \(\large\frac{\mathrm{d}y}{\mathrm{d}x}\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\sin x)=\cos x\\ \large\frac{\mathrm{d}}{\mathrm{d}x}(\cos x)=-\sin x\)
If \(\large f(x) = 3\sin x +a \cos x\) , and \(\large f'(x)=b\cos x-2 \sin x\)
Find a and b
a =
b =
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(3\sin x)=3\cos x\\ \large\frac{\mathrm{d}}{\mathrm{d}x}(2\cos x)=-2\sin x\)
If \(\large\frac{\mathrm{d}}{\mathrm{d}x}(a\sin x+b \cos x)=5 \sin x-3 \cos x\)
Find a and b
a =
b =
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(-5\cos x)=5\sin x\\ \large\frac{\mathrm{d}}{\mathrm{d}x}(-3\sin x)=-3\cos x\)
If \(\large f(x) = 2\sin x \) , find \(\large f'(\frac{\pi}{3})\)
\(\large f'(\frac{\pi}{3})\) =
\(\large f'(x) = 2\cos x \\ \large f'(\frac{\pi}{3}) = 2\cos (\frac{\pi}{3}) =2\times\frac{1}{2}=1\)
If \(\large f(x) = 2\cos x \) , and \(\large f'(\frac{\pi}{a})=-\sqrt{2}\)
find a
a =
\(\large f'(x) = -2\sin x \\ \large f'(\frac{\pi}{a}) = -2\sin (\frac{\pi}{a})=-\sqrt{2}\)
\(\large\sin (\frac{\pi}{a})=\frac{\sqrt{2}}{2}\)
\(\large\sin (\frac{\pi}{4})=\frac{\sqrt{2}}{2}\)
Work out \(\large\frac{\mathrm{d}}{\mathrm{d}x}(x^2+e^x)\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(e^x)=e^x\)
If \(\large f(x) = \frac{\ln x}{2}\) , work out \(\large f'(x)\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)=\frac{1}{x}\)
Work out \(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln 2x)\)
Using the laws of logarithms, we know that
\(\large\ln 2x=\ln 2+\ln x \)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln 2+\ln x)=0+\frac{1}{x}\)
\(\large y=\frac{2e^x+1}{3}\)
Find \(\large\frac{\mathrm{d}y}{\mathrm{d}x}\) when x = 0
\(\large y=\frac{2e^x+1}{3}=\frac{2}{3}e^x+\frac{1}{3}\)
\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{3}e^x\)
when x = 0 , \(\large\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{3}e^0=\frac{2}{3} \times 1=\frac{2}{3}\)
\(\large f(x) = 3e^x-2 \sin x\) , work out \(\large f'(0)\)
\(\large f'(0)\) =
\(\large f'(x) = 3e^x-2 \cos x\)
\(\large f'(0) = 3e^0-2 \cos 0=3\times 1-2\times 1=1\)
Here's a quiz that mainly practises the Chain Rule
START QUIZ!
Mixed SL Differentiation 1 1/1
Find the derivative of (2x - 5)4
We can use the chain rule
y = u4 | u = 2x - 5 |
\(\frac { dy }{ du } \) = 4u3 | \(\frac { du }{ dx } \) = 2 |
\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)
= 8u3
=8(2x - 5)3
Find the derivative of e5x
You can use the chain rule, but you should try to remember that \(\frac{d}{dx}(e^{ax})=ae^{ax}\)
Find the gradient function of f(x) = ln3x
There are 2 ways of thinking about this type of derivative.
- You can use the chain rule \(f'(x)=\frac{3}{3x}=\frac{1}{x}\)
or
- You can use the properties of logarithms
f(x) = ln3x = ln3 + lnx
\(f'(x)=0+\frac{1}{x}\)
Find \(\frac { dy }{ dx } \) if \(y =\frac{1}{e^{x^{3}}}\)
Notice that we can re-write the question to make it easier
\(y =\frac{1}{e^{x^{3}}}=e^{-x^{3}}\)
We can use the chain rule
y = eu | u = \(-x^{3}\) |
\(\frac { dy }{ du } \) = eu | \(\frac { du }{ dx } \) = -3x² |
\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)
= eu(-3x²)
=\(-3x^{2}e^{-x^{3}}\)
Find the derivative of \( {1\over 4x}\)
It helps if you think of the question like this
\( {1\over 4x}= {1\over 4} x^{-1}\)
Some people get integration and differentiation mixed up for logarithm functions
Note that \(\int { \frac { 1 }{ 4x } dx= } \frac { 1 }{ 4 } \int { \frac { 1 }{ x } dx= } \frac { 1 }{ 4 } lnx \)
Find \(\frac { d }{ dx } (lnx)^{3}\)
Use the chain for this, or use the quick rule
\(\frac { d }{ dx } [f(x)]^{ n }=nf'(x)[f(x)]^{ n-1 }\)
\(\frac { d }{ dx }(lnx)^{3}=3{1\over x}(lnx)^{2}\)
Find \(\frac { d }{ dx } (lnx^{3})\)
Use laws of logarithms to transform this question into something easier
\(lnx^{3}=3lnx\)
Differentiate sin2x
You should try to remember that \(\frac { d }{ dx } (sinax)=acosax\)
Find \(\frac { d }{ dx } (sinx^{2})\)
We can use the chain rule
y = sinu | u = x² |
\(\frac { dy }{ du } \) = cosu | \(\frac { du }{ dx } \) = 2x |
\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)
= cosu(2x)
= \(2x(cosx^{2})\)
Differentiate \(\sqrt{sinx}\)
We can use the chain rule
y = \(u^{\frac{1}{2}}\) | u = sinx |
\(\frac { dy }{ du } \) = \(\frac{1}{2}u^{-\frac{1}{2}}\) | \(\frac { du }{ dx } \) = cosx |
\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)
= \(\frac{1}{2}u^{-\frac{1}{2}}cosx\)
= \(\frac{1}{2}(sinx)^{-\frac{1}{2}}cosx\)
=\(\frac{cosx}{2\sqrt{sinx}}\)
Here's a quiz that mainly practises the Chain Rule and the Product and Quotient Rule
START QUIZ!
Mixed SL Differentiation 2 1/1
Find \(\frac { d }{ dx } ({ e }^{ \pi -x })\)
We can use the chain rule
y = eu | u = \(\pi-x\) |
\(\frac { dy }{ du } \) = eu | \(\frac { du }{ dx } \) = -1 |
\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)
= eu(-1)
=\(-{ e }^{ \pi -x }\)
Work out \(\frac { d }{ dx } [cos(2x-\frac { \pi }{ 4 } )]\)
y = cosu | u = \(2x-\frac { \pi }{ 4 } \) |
\(\frac { dy }{ du } \) = -sinu | \(\frac { du }{ dx } \)=2 |
\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)
=-sinu (2)
=\(-2sin(2x-\frac { \pi }{ 4 } )\)
\(f(x) = tan(\pi -2x)\)
Work out \(f'(x)\)
We can use the chain rule
y = tanu | u = \( { \pi }-2x\) |
\(\frac { dy }{ du } \) = sec²u | \(\frac { du }{ dx } \)=-2 |
\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)
=sec²u (-2)
=\(-2{ sec }^{ 2 }(\pi -2x)\)
Differentiate \(3x(2x-1)^5\)
We need to use the product rule
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
u = 3x | v = \((2x-1)^5\) |
\(\frac { du }{ dx } \) = 3 | \(\frac { dv }{ dx } =10(2x-1)^4\) |
\( \frac{dy}{dx}=3x\times10(2x-1)^{4} \ +\ (2x-1)^{5}\times 3\)
We can factorise this result. There is a common factor of \(3(2x-1)^4\)
\( \frac{dy}{dx}=3(2x-1)^{4}[10x+(2x-1)]\)
\( \frac{dy}{dx}=3(2x-1)^{4}(12x-1)\)
Differentiate sinxcos2x
We need to use the product rule
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
u = sinx | v = cos2x |
\(\frac { du }{ dx } \) = cosx | \(\frac { dv }{ dx } =-2sin2x\) |
\( \frac{dy}{dx}=sinx\times(-2sin2x) \ +cosx\times cos2x\)
Differentiate \(x^{2}lnx\)
We need to use the product rule
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
u = x² | v = lnx |
\(\frac { du }{ dx } \) = 2x | \(\frac { dv }{ dx } ={1 \over x}\) |
\( \frac{dy}{dx}=x^2 \times{1 \over x} \ +lnx\times 2x\)
= x + 2xlnx
Work out \(\frac { d }{ dx } (2x\sqrt { 1-{ x }^{ 2 } } )\)
We need to use the product rule
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
u = 2x | v = \(\sqrt{1-x^2}=(1-x^2)^{1 \over 2}\) |
\(\frac { du }{ dx } \) = 2 | \(\frac { dv }{ dx } ={1 \over 2}(-2x)(1-x^2)^{-\frac{1}{2}}=\frac{-x}{\sqrt{1-x^2}}\) |
\( \frac{dy}{dx}=2x \times\frac{-x}{\sqrt{1-x^2}} \ +\sqrt{1-x^2}\times 2\)
\(=\frac{-2x^2}{\sqrt{1-x^2}} \ +2\sqrt{1-x^2}\)
\(=\frac{-2x^2}{\sqrt{1-x^2}} \ + \ \frac{2(1-x^2)}{\sqrt{1-x^2}}\)
\(=\frac{2-4x^2}{\sqrt{1-x^2}} \)
Differentiate \(\frac{e^x}{sinx}\)
We need to use the quotient rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
\(u = e^x\) | v = sinx |
\(\frac{du}{dx}=e^x\) | \(\frac{dv}{dx}=cosx\) |
\(\frac { dy }{ dx } =\frac { sinx\cdot e^x -e^x\cdot cosx }{( sinx)^{ 2 }} \)
\(=\frac { e^x(sinx - cosx) }{( sinx)^{ 2 }} \)
Work out \(\frac{d}{dx}(\frac{e^{3x}}{cos2x})\)
We need to use the quotient rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
\(u=e^{3x}\) | v = cos2x |
\(\frac{du}{dx}=3e^{3x}\) | \(\frac{dv}{dx}=-2sin2x\) |
\(\frac { dy }{ dx } =\frac { cos2x\cdot 3e^{3x} -e^{3x}\cdot (-2sin2x) }{ (cos2x)^{ 2 } }\)
\(\frac { dy }{ dx } =\frac { 3e^{3x}cos2x +2e^{3x}sin2x }{ (cos2x)^{ 2 } }\)
Differentiate \(\frac{x^2+1}{3x-2}\)
We need to use the quotient rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = x² + 1 | v = 3x - 2 |
\(\frac{du}{dx}=2x\) | \(\frac{dv}{dx}=3\) |
\(\frac { dy }{ dx } =\frac { (3x-2)\cdot (2x) -(x^2+1)\cdot 3}{ (3x-2)^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { 6x^2-4x -3x^2-3}{ (3x-2)^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { 3x^2-4x -3}{ (3x-2)^{ 2 } } \)
Here's a quiz that mainly practises the Product and Quotient Rule
START QUIZ!
Mixed SL Differentiation 3 1/1
Differentiate \(\frac{2x}{\sqrt{1-x^2}}\)
We need to use the Quotient Rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = 2x | \(v=\sqrt{1-x^2}\\v=(1-x^2)^{0.5}\) |
\(\frac { du }{ dx } =2\) | \(\frac { dv }{ dx } =0.5(-2x)(1-x^2)^{-0.5}\) \(\frac { dv }{ dx } =\frac{-x}{\sqrt{1-x^2}}\) |
\(\frac { dy }{ dx } =\frac { \sqrt{1-x^2}\cdot 2-2x\cdot \frac{-x}{\sqrt{1-x^2}} }{ 1-x^2 }\)
\(\frac { dy }{ dx } =\frac { \frac{2(1-x^2)}{\sqrt{1-x^2}}+ \frac{2x^2}{\sqrt{1-x^2}} }{ 1-x^2 }\)
\(\frac { dy }{ dx } =\frac { \frac{2-2x^2+2x^2}{\sqrt{1-x^2}}}{ 1-x^2 }\)
\(\frac { dy }{ dx } =\frac{2}{ (1-x^2)^{\frac{3}{2}} }\)
Differentiate \(e^{2x}lnx\)
We need to use the product rule
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
\(u=e^{2x}\) | v=lnx |
\(\frac { du }{ dx } =2e^{2x}\) | \(\frac { dv }{ dx } =\frac{1}{x}\) |
\(\frac { dy }{ dx } =e^{ 2x }\frac { 1 }{ x } +lnx\cdot 2e^{ 2x }\)
\(\frac { dy }{ dx } =e^{ 2x }(\frac { 1 }{ x } +2lnx)\)
Differentiate \(e^xlnx^2\)
We need to use the product rule
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
\(u=e^{x}\) | v=lnx² v=2lnx |
\(\frac { du }{ dx } =e^{x}\) | \(\frac { dv }{ dx } =2\cdot\frac{1}{x}\) |
\(\frac { dy }{ dx } =e^{ x }\frac { 2 }{ x } + 2e^{ x }lnx\)
\(\frac { dy }{ dx } =\frac { 2e^x }{ x } + 2e^{ x }lnx\)
Differentiate \(\frac { e^{ { 2x } } }{ lnx } \)
We need to use the Quotient Rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = \(e^{2x}\) | v = lnx |
\(\frac { du }{ dx } =2e^{2x}\) | \(\frac { dv }{ dx } =\frac{1}{x}\) |
\(\frac { dy }{ dx } =\frac { lnx\cdot 2e^{ 2x }-e^{ 2x }\cdot \frac { 1 }{ x } }{ (lnx)^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { e^{ 2x }(2lnx -\frac { 1 }{ x }) }{ (lnx)^{ 2 } } \)
Differentiate \(\frac { { e }^{ 3x } }{ ln2x } \)
We need to use the Quotient Rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = \(e^{3x}\) | v = ln2x |
\(\frac { du }{ dx } =3e^{3x}\) | \(\frac { dv }{ dx } =\frac{1}{x}\) |
\(\frac { dy }{ dx } =\frac { ln2x\cdot 3e^{ 3x }-e^{ 3x }\cdot \frac { 1 }{ x } }{ (ln2x)^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { e^{ 3x }(3ln2x-\frac { 1 }{ x } ) }{ (ln2x)^{ 2 } } \)
Differentiate \(\frac { ln3x }{ { e }^{ 2x } } \)
We need to use the Quotient Rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = ln3x | v = \(e^{2x}\) |
\(\frac { du }{ dx } =\frac{1}{x}\) | \(\frac { dv }{ dx } =2e^{2x}\) |
\(\frac { dy }{ dx } =\frac { e^{ 2x }\cdot \frac { 1 }{ x } -2e^{ 2x }\cdot ln3x }{ ({ e }^{ 2x })^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { e^{ 2x }(\frac { 1 }{ x } -2ln3x) }{ ({ e }^{ 2x })^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { \frac { 1 }{ x } -2ln3x }{ { e }^{ 2x } } \)
\(\frac { dy }{ dx } =\frac { \frac { 1 }{ x } -\frac { 2xln3x }{ x } }{ { e }^{ 2x } } \)
\(\frac { dy }{ dx } =\frac { 1-2xln3x }{ { xe }^{ 2x } } \)
Differentiate \(\frac { { e }^{ x } }{ tanx } \)
We need to use the Quotient Rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = \(e^{x}\) | v = tanx |
\(\frac { du }{ dx } =e^{x}\) | \(\frac { dv }{ dx } =sec²x\) |
\(\frac { dy }{ dx } =\frac { { tanx }\cdot e^{ x }-e^{ x }sec²x }{ (tanx)^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { { e^xtanx }-e^{ x }sec²x }{ (tanx)^{ 2 } } \)
If \(f(x) = 2\sqrt{x}\) , then \(f''(x)=ax^b\)
Work out a and b
a =
b =
\(f(x)=2x^{0.5}\)
\(f'(x)=2(0.5)x^{-0.5}=x^{-0.5}\)
\(f''(x)=-0.5x^{-1.5}\)
If \(f(x)=\frac { { x }^{ 3 } }{ { x }^{ 2 }+1 } \), find the gradient of the function when x = -1
\(f'(-1)\) =
We need to use the Quotient Rule
\(f(x)=\frac { g(x) }{ h(x) } \\ f'(x)=\frac { h(x)\cdot g'(x)-g(x)\cdot h'(x) }{ { \left[ h(x) \right] }^{ 2 } } \)
g(x) = \(x^3\) | h(x) = x²+1 |
g'(x) = 3x² | h'(x) = 2x |
\(f'(x)=\frac { (x²+1)(3x²)-{ x }^{ 3 }(2x) }{ { (x²+1) }^{ 2 } } \)
\(f'(-1)=\frac { ((-1)²+1)(3(-1)²)-{ (-1) }^{ 3 }(2(-1)) }{ {((-1)²+1 ) }^{ 2 } } \)
\(f'(-1)=\frac { (2)(3)-{ (-1) }(-2) }{ { (2) }^{ 2 } } \)
\(f'(-1)=\frac { 4 }{ 4 } \)
The graph of the function \(f(x)=\frac{lnx}{x}\), x>0 has a maximum value at x = a
Work out the value of a
a =
The local maximum of a function occurs when the gradient = 0
We need to use the Quotient Rule to find the gradient function
\(f(x)=\frac { g(x) }{ h(x) } \\ f'(x)=\frac { h(x)\cdot g'(x)-g(x)\cdot h'(x) }{ { \left[ h(x) \right] }^{ 2 } } \)
g(x) = lnx | h(x) = x |
\(g'(x) =\frac{1}{x}\) | h'(x) = 1 |
\(f'(x) =\frac { { x }\cdot \frac { 1 }{ x } -lnx }{ x² } \)
\(f'(x)=\frac { 1-lnx }{ x² } \)
Solve \(f'(x) = 0\)
\(\frac { 1-lnx }{ x² } =0\)
1 - lnx = 0 , since
lnx = 1
x = e
Here's a quiz that mainly practises the Related Rates of Change
START QUIZ!
Related Rates of Change 1/1
The rate at which the area, A, of a square of length x is increasing is given by \(\large \frac{dA}{dt}\)
Which of the following will find the rate of increase in the length of the side, \(\large \frac{dx}{dt}\)
This is a related rates of change question. We use the chain rule
\(\large \frac{dA}{dt}=\frac{dA}{dx} \cdot \frac{dx}{dt}\)
The area of the square is given by the formula
\(\large A=x^2\)
We differentiate with respect to x to find \(\large \frac{dA}{dx}\)
\(\large \frac{dA}{dx}=2x\)
We substitute this into the chain rule and rearrange
\(\large \frac{dA}{dt}=2x \cdot \frac{dx}{dt}\)
\(\large \frac{\frac{dA}{dt}}{2x}= \frac{dx}{dt}\)
The rate at which the radius, r, of a sphere is increasing is given by \(\large \frac{dr}{dt}\)
Which of the following will find the rate of increase of the volume, \(\large \frac{dV}{dt}\)
The volume of a sphere is given by \(\large V= \frac{4}{3}\pi r^3 \)
This is a related rates of change question. We use the chain rule
\(\large \frac{dV}{dt}=\frac{dV}{dr} \cdot \frac{dr}{dt}\)
The volume of a sphere is given by
\(\large V= \frac{4}{3}\pi r^3 \)
We differentiate with respect to r to find \(\large \frac{dV}{dr}\)
\(\large \frac{dV}{dr}=4\pi r^2\)
We substitute this into the chain rule
\(\large \frac{dV}{dt}=4\pi r^2 \cdot \frac{dr}{dt}\)
The radius, r, of a circle is increasing at the rate of \(\large \frac{3}{\pi} \ cms^{-1}\)
Find the rate at which the circumference, C, is increasing
\(\large \frac{dC}{dt}\)= \(\large cms^{-1}\)
This is a related rates of change question. We use the chain rule
\(\large \frac{dC}{dt}=\frac{dC}{dr} \cdot \frac{dr}{dt}\)
The circumference of a circle is given by the formula
\(\large C=2 \pi r\)
We differentiate with respect to r to find \(\large \frac{dC}{dr}\)
\(\large \frac{dC}{dr}=2\pi\)
We substitute this into the chain rule
\(\large \frac{dC}{dt}=2\pi \cdot \frac{3}{\pi} =6\)
The radius, r, of a circle is increasing at the rate of \(\large \frac{1}{\pi} \ cms^{-1}\)
Find the rate at which the circumference, A, is increasing when r = 5cm
\(\large \frac{dA}{dt}\)= \(\large cm^2s^{-1}\)
This is a related rates of change question. We use the chain rule
\(\large \frac{dA}{dt}=\frac{dA}{dr} \cdot \frac{dr}{dt}\)
The area of a circle is given by the formula
\(\large A=\pi r^2\)
We differentiate with respect to r to find \(\large \frac{dA}{dr}\)
\(\large \frac{dA}{dr}=2\pi r\)
When r = 5cm, \(\large \frac{dA}{dr}=10\pi \)
We substitute this into the chain rule
\(\large \frac{dA}{dt}=10\pi \cdot \frac{1}{\pi} =10\)
The volume of a cube is increasing at a rate of 30\(cm^3s^{-1}\)
Find the rate at which the sides of the cube, x, are increasing when the lengths of the sides are 10cm
\(\large \frac{dx}{dt}\)= \(\large cms^{-1}\)
This is a related rates of change question. We use the chain rule
\(\large \frac{dV}{dt}=\frac{dV}{dx} \cdot \frac{dx}{dt}\)
The volume of the cube is given by the formula
\(\large V=x^3\)
We differentiate with respect to x to find \(\large \frac{dV}{dx}\)
\(\large \frac{dV}{dx}=3x^2\)
When x = 10, \(\large \frac{dV}{dx}=300\)
We substitute this into the chain rule
\(\large30=300 \cdot \frac{dx}{dt}\)
\(\large \frac{dx}{dt}\)= 0.1
A spherical balloon is deflating at a rate of \(\large 160\pi \ cm^3s^{-1}\)
At what rate is the radius of the balloon decreasing when the radius, r = 4cm ?
The volume of a sphere is given by \(\large V= \frac{4}{3}\pi r^3 \)
\(\large \frac{dr}{dt}\)= \(\large cms^{-1}\)
This is a related rates of change question. We use the chain rule
\(\large \frac{dV}{dt}=\frac{dV}{dr} \cdot \frac{dr}{dt}\)
The volume of a sphere is given by \(\large V= \frac{4}{3}\pi r^3 \)
We differentiate with respect to r to find \(\large \frac{dV}{dr}\)
\(\large \frac{dV}{dr}=4\pi r^2\)
when the radius, r = 4cm , \(\large \frac{dV}{dr}=4\pi \times 4^2=64 \pi\)
We substitute this into the chain rule
\(\large160\pi=64\pi\cdot \frac{dr}{dt}\)
\(\large \frac{dr}{dt}=\frac{160\pi}{64\pi}=2.5\)
Here's a quiz that practises differentiating using a graphical display calculator
START QUIZ!
Using GDC for Differentiation 1/1
Use your graphic display calculator to solve this question
Let \(\large f(x)=3^{-x}+4x-5\)
The graph of f has a horizontal tangent at the point where x = a
Find the value of a
Give your answer to 3 s.f.
a =
A horizontal tangent has a gradient = 0
You need to find the value of x for which \(\large f'(x)=0\)
Use your graphic display calculator to solve this question
Let \(\large f(x)=\ln 2x +2^x\)
The tangent to the graph of f at x=1 intersects the y axis at y = a
Find the value of a
Give your answer to 3 s.f.
a =
Use your graphical display calculator to sketch the graph of the tangent to the function f at x = 1
The equation of the tangent is found to be y = 2.39x + 0.307
Hence this intersects the y axis when x = 0
To 3 significant figures, this is y = 0.307
Use your graphic display calculator to solve this question
Let \(\large f(x)=\frac{2x}{2x-5}\ , \ x\neq\frac{5}{2}\)
The normal to the graph of f at x = 0 intersects the graph of f again at x = a
Find a
Give the exact answer
a =
Use your graphical display calculator to sketch the graph of the normal to the function f at x = 0
Plot f and the equation of the normal on the same axes and find the point of intersection
Use your graphic display calculator to solve this question
Let \(\large f'(x)=\sin2 x\cos3 x\ ,\ 0\le x\le1\)
The graph of f has two points of inflexion at x = a and x = b, where a < b
Find the value of a
Give your answer to 3 s.f.
a =
Make sure that your calculator is in radians mode
Notice that the domain for \(\large f'\) exists for \(0\le x\le1\)
The points of inflexion for the graph of f exist where \(\large f''(x)=0\)
You can plot the graph of \(\large f'\) and find the x coordinate of the first stationary point for \(\large f'\), in this case, a local maximum
Use your graphic display calculator to solve this question
Let \(\large f(x)=x^4-2x^2+x\)
The following diagram shows the graph of f
Find the x coordinates of the point A
x =
Use your calculator to find the root of the equation f(x) = 0
Ensure that you find the correct root
Use your graphic display calculator to solve this question
Let \(\large f(x)=\ln (x)-3x\ , \ x>0\)
Solve \(\large f'(x)=f''(x)\)
Give your answer to 3 s.f.
x =
\(\large f(x)=\ln (x)-3x\ , \ x>0\)
\(\large f'(x)=\frac{1}{x}-3\)
\(\large f''(x)=-\frac{1}{x^2}\)
Solve \(\large \frac{1}{x}-3=-\frac{1}{x^2}\)
You can do this by plotting the graphs of \(\large y=\frac{1}{x}-3\) and \(\large y=-\frac{1}{x^2}\) and finding the x coordinate of the point of intersection
Use your graphic display calculator to solve this question
Let \(\large f(x)=x^3e^{-2x}\)
Find the x coordinate of the point of intersection of the tangents to the graph of f at x = 1 and x = 2
Give your answer to 2 s.f.
x =
Note that the answer only needs to be given to 2 s.f.
Use your graphical display calculator to sketch the graph of the tangent to the function f at x = 1 and also for x = 2
Plot these two tangents and find the point of intersection
Use your graphic display calculator to solve this question
Let \(\large f(x)=x^4-2e^x\)
The following diagram shows the graph of f
A is a point of inflexion
The coordinates of A are (a,f(a))
Find the value of a
Give your answer to 3 s.f.
a =
To find a point of inflexion, we need to solve \(\large f''(x)=0\)
This is also a stationary point of the function \(\large f'(x)\)
We can differentiate the function to find \(\large f'(x)=4x^3-2e^x\)
Use your graphic display calculator to solve this question
Let \(\large f(x)=sin(e^x) \ , \ 0\le x \le 2\)
The curve of f is concave up on the interval a < x < b
Find the values of a and b
Give your answer to 3 s.f.
a =
b =
The curve of a function is concave up when \(\large f''(x)>0\)
This will occur between two points of inflexion
Differentiate \(\large f(x)=sin(e^x) \) using the Chain Rule
\(\large f'(x)=e^xcos(e^x) \)
We can find the points of inflexion for \(f(x)\) by finding the stationary points for \(f'(x)\)
Use your graphic display calculator to solve this question
Let \(\large f(x)=2x \ln x \ ,\ x>0\)
Points P(1,0) and Q are on the curve of f.
The tangent to the curve of f at P is perpendicular to the tangent to the curve at Q.
Find the x coordinates of Q.
Give your answer to 3 s.f.
a =
You do not need to plot a graph of the curve and the tangents, but it may help you visualise the problem.
Plot the graph of the function and the tangent to the curve at the point x = 1
To get a true picture of the graph, you may want to plot the graph so that the axes are 1:1 (so that a right angle actually looks like a right angle).
You will notice that the gradient of this tangent = 2
You need to find the x coordinates of the graph for which the curve has a gradient = \(-\frac{1}{2}\)
Differentiate the function using the product rule
\(\large f'(x)=2 \ln x+2 \)
Solve \(\large 2 \ln x+2 =-\frac{1}{2}\)
You can use the equation solver or a graphical approach.
Here is what the two tangents look like
Here's a quiz that practises Graphs and the Gradient Function
START QUIZ!
Mixed SL Differentiation 4 1/1
The diagram shows the graph of y = f(x)
Which of the following is true?
Select ALL correct answers
For a local maxima
- \(f'(x)=0\)
- \(f''(x)<0\)
The diagram shows the graph of y = f(x)
Which of the following is the correct graph of \(y = f'(x)\)
A | |
B | |
C | |
D |
The diagram shows the graph of y = f(x)
Which of the following is true
Select ALL correct answers
At local maxima \(f'(x)\)= 0 and \(f''(x)<0\)
At local minima \(f'(x)\)= 0 and \(f''(x)>0\)
Note that, gradient at c is negative and gradient at e is positive. Therefore, \(f'(c)
The diagram shows the graph of \(y=f'(x)\)
Which of the following is TRUE about \(y=f(x)\)
Select ALL correct answers
The diagram below shows the graph of the function y = f(x) in black
The diagram shows the graph of \(y=f'(x)\)
Which of the following is TRUE
Select ALL correct answers
The diagram below shows the graph of the function \(y = f''(x)\) in green
The diagram shows the graph of \(y=f'(x)\)
Which is the correct sketch of the graph of \(y=f(x)\)
A | |
B | |
C | |
D |
We are given the graph of the gradient function and asked to work out what the original function might look like
The function f(x) has a local maxima at x = b.
a is a value close to b such that a<b
Which of the following is true?
Select ALL correct answers
The function f(x) has a local minima at x = b.
a and c are values close to b such that a<b<c
Which of the following is true?
Select ALL correct answers
The function f has a non-stationary point of inflexion at x = b
a and c are values close to b such that a < b < c
Which of the following is true.
Select ALL correct answers.
<>
If there is a point of inflexion then \(f''(x)=0\)
If it is a non-stationary point then \(f'(x)\neq 0\)
The function must either be
- increasing from x=a to x=c, so \(f(a)<f(b)<f(c)\)<>
<> - decreasing from x=a to x=c, so \(f(a)>f(b)>f(c)\)
The sign of the gradient at x=a and x=c must be the same (they must be either both positive, or both negative).
The following is true about a function f
- <>
<><><> a < b < c<><> - f(x) is increasing on the interval [a,c]
- \(f'(x)<0\) for \(x < a\)
- \(f''(b)=0\)
- \(f'(c)=0\)
Which of the following is TRUE?
Here's a quiz that practises Stationary Points and Equations of Tangents and Normals
START QUIZ!
Mixed SL Differentiation 5 1/1
Find the gradient of the curve \(y = x^3-2x^2\) where x= 1
gradient =
Differentiate \(y = x^3-2x^2\)
\(\frac{dy}{dx}=3x^2-4x\)
when x = 1, \(\frac{dy}{dx}=3(1)^2-4(1)=-1\)
The gradient of the curve y = sin3x at the point A is 1.5
The x cordinate of A is \(\frac{\pi}{k}\)such that \(0<\frac{\pi}{k}<\frac{\pi}{2}\)
Work out the value of k
k =
y = sin3x,
\(\frac{dy}{dx}=3cos3x\)
We know that \(\frac{dy}{dx}=\frac{3}{2}\)
\(3cos3x=\frac{3}{2}\)
\(cos3x=\frac{1}{2}\)
Arccos(0.5)=\(\frac{\pi}{3}\)
3x = \(\frac{\pi}{3}\)
x = \(\frac{\pi}{9}\)
What is the gradient of the normal to the curve \(y = \frac{e^x}{x^2}\) at x = 1
Differentiate \(y = \frac{e^x}{x^2}\)
We need to use the Quotient Rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = \(e^x\) | v = x² |
\(\frac{du}{dx}=e^x\) | \(\frac{dv}{dx}=2x\) |
\(\frac{dy}{dx}=\frac { x^{ 2 }e^{ x }-e^{ x }\cdot 2x }{ (x^{ 2 })^{ 2 } } \)
when x = 1 , \(\frac{dy}{dx}=\frac { 1^{ 2 }e^{ 1 }-2e^{ 1 } }{ 1^{ 4 } } \)
\(\frac{dy}{dx}=\frac { e-2e }{ 1 } =-e\)
Gradient of tangent = -e
Gradient of normal = \(\frac{-1}{-e}=\frac{1}{e}\)
Find the gradient of the tangent to \(f(x)=\frac{1}{e^x}\) at x=ln2
gradient =
\(f(x)=e^{-x}\)
\(f'(x)=-e^{-x}\)
\(f'(x)= -\frac{1}{e^x}\)
\(f'(ln2)= -\frac{1}{e^{ln2}}=-\frac{1}{2}\)
Find the equation of the tangent to the curve y = xtanx at \((\pi,0)\)
To differentiate y = xtanx, we need to use the Product Rule
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
u = x | v = tanx |
\(\frac{du}{dx}=1\) | \(\frac{dv}{dx}=sec²x\) |
\(\frac{dy}{dx}=xsec²x+tanx\cdot1\)
When x = \(\pi\), \(\frac{dy}{dx}=\pi sec²(\pi)+tan(\pi)\)
\(\frac{dy}{dx}=\frac{\pi}{[ cosx(\pi)]^2}+tan(\pi)\)
\(\frac{dy}{dx}=\frac{\pi}{( -1)^2}+0=\pi\)
Find the equation of the straight line with gradient = \(\pi\) that passes through the point \((\pi,0)\)
\(\frac{y-0}{x-\pi}=\pi\)
\(y=\pi(x-\pi)\)
\(y=\pi x-\pi^2\)
Find the equation of the normal to the curve \(y=x^3lnx\) at (1,0)
We need to use the product rule to differentiate \(y=x^3lnx\)
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
u = \(x^3\) | v = lnx |
\(\frac{du}{dx}=3x^2\) | \(\frac{dv}{dx}=\frac{1}{x}\) |
\(\frac{dy}{dx}=x^3\cdot \frac{1}{x}+lnx\cdot3x^2\)
\(\frac{dy}{dx}=x^2+3x^2lnx\)
\(\frac{dy}{dx}=x^2(1+lnx)\)
When x = 1, \(\frac{dy}{dx}=1(1+ln1)=1(1+0)=1\)
Gradient of tangent = 1
Gradient of normal = -1
Find the equation of the straight line with gradient = -1 which passes through the point (1,0)
\(\frac{y-0}{x-1}=-1\)
y = -x + 1
Find the equation of the normal to the curve \(y=e^{sin2x}\) at \((\frac{\pi}{2},1)\)
\(y=e^{sin2x}\)
\(\frac{dy}{dx}=2cos2x\cdot e^{sin2x}\)
When x = \(\frac{\pi}{2}\), \(\frac{dy}{dx}=2cos(\pi)\cdot e^{sin(\pi)}\)
\(\frac{dy}{dx}=2(-1))\cdot e^{0}=-2\)
Gradient of tangent = -2
Gradient of normal = \(\frac{-1}{-2}=\frac{1}{2}\)
Find the equation of the straight line with gradient = \(\frac{1}{2}\) which passes through the point \((\frac{\pi}{2},1)\)
\(\frac{y-1}{x-\frac{\pi}{2}}=\frac{1}{2}\)
\(2y-2=x-\frac{\pi}{2}\)
\(4y-4=2x-\pi\)
\(2x-4y+(4-\pi)=0\)
The curve \(y=\frac{lnx}{x}\) has a maximum value at the point A.
Find the x coordinate of A
x coordinate of A =
Find the gradient function using the Quotient Rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = lnx | v = x |
\(\frac{du}{dx}=\frac{1}{x}\) | \(\frac{dv}{dx}=1\) |
\(\frac{dy}{dx}=\frac { x\cdot\frac{1}{x}-lnx\cdot 1 }{ x^{ 2 } } \)
\(\frac{dy}{dx}=\frac { 1-lnx }{ x^{ 2 } } \)
Maximum value occurs where \(\frac{dy}{dx}=0 \)
Solve \(\frac { 1-lnx }{ x^{ 2 } } =0\)
Since \(x\neq 0\) , 1-lnx = 0
lnx = 1
x = e
The function f(x) = tanx - 4x, \(0 <x<\frac { \pi }{ 2 } \) has a minimum value at \(x = \frac { \pi }{ a} \)
Work out the value of a
a =
Differentiate the function f(x) = tanx - 4x
\(f'(x)=sec^2x-4\)
Minimum value occurs when \(f'(x)=0\)
sec²x - 4 = 0
sec²x = 4
cos²x = \(\frac{1}{4}\)
cosx = \(\pm \frac { 1 }{ 2 } \)
\(Arccos(\frac{1}{2})=\frac{\pi}{3}\)
There is only one correct value in the interval \(0 <x<\frac { \pi }{ 2 } \)
Hence \(x=\frac{\pi}{3}\)
Find the x coordinate of the point of inflexion on the curve \(y =xe^x\)
x =
Use the Product Rule to find \(\frac{dy}{dx}\)
\(\frac{dy}{dx}=e^x+xe^x\)
Differentiate again to find \(\frac{d²y}{dx²}\)
\(\frac{d²y}{dx²}=e^x+e^x+xe^x\)
\(\frac{d²y}{dx²}=e^x(2+x)\)
Point of inflexion exists where \(\frac{d²y}{dx²}=0\)
\(e^x(2+x)=0\)
\(e^x=0\quad,\quad2+x=0\)
not possible , x = -2
Here's a quiz that practises Kinematics
START QUIZ!
Mixed SL Differentiation 6 1/1
An object is moving along a line such that its displacement, s at any point in time t is given by
s(t) = 3sint - 4cost
What is the object's initial displacement?
Initial displacement =
The initial displacement is given by s(0)
s(0) = 3sin(0) - 4cos(0)
s(0) = -4(1) = -4
An object moves in a straight line with position relative to the origin O given by
s(t) = sin4t - cos3t
where t is the time in seconds.
Which expression describes the velocity of the object?
Velocity is given by the rate of change of displacement. We need to differentiate displacement:
s(t) = sin4t - cos3t
\(v(t)=s'(t) = 4cos4t + 3sin3t\)
The velocity of a particle moving in a straight line is given by
\(v(t) = 10e^{-0.5t}\)
What is the initial acceleration of the particle?
Initial acceleration =
Acceleration is the rate of change of velocity. We need to differentiate velocity
\(v(t) = 10e^{-0.5t}\)
\(a(t) = v'(t) = 10(-0.5)e^{-0.5t}\)
\(a(t) = -5e^{-0.5t}\)
Initial acceleration is a(0)
\(a(0) = -5e^{-0.5(0)}\)
\(a(0) = -5e^{0}=-5\)
The displacement, s of an object is given by
s(t) = 1 - cos2t
where t is time.
The first time that the object's velocity is a maximum is when t = \(\frac{\pi}{b}\)
Find b
b =
Velocity is given by the rate of change of displacement. We need to differentiate displacement:
s(t) = 1 - cos2t
\(v(t)=s'(t) = 2sin2t\)
Maximum velocity occurs when \(v'(t)=0\)
\(v'(t) = 4cos2t\)
Solve 4cos2t = 0
cos2t = 0
Arccos(0) = \(\frac{\pi}{2}\)
\(2t=\frac{\pi}{2}\)
\(t=\frac{\pi}{4}\)
The displacement, s of an object is given by \(s(t) = e^{-2t}\), where t is time.
Which is the correct expression for acceleration?
Acceleration is the rate of change of velocity.
Velocity is given by the rate of change of displacement.
We need to differentiate two times:
\(s(t) = e^{-2t}\)
\(v(t)=s'(t) = -2e^{-2t}\)
\(a(t)=v'(t)=s''(t) = 4e^{-2t}\)
A ball is thrown vertically upwards. The height, s of the ball after time t is given by the equation
\(s(t) = bt - 5t^2\)
If the ball reaches its maximum height when t = 2.5, find the value of b
b =
The ball reaches its maximum height when velocity = 0
To find velocity we need to differentiate the equation for displacement (height):
\(s(t) = bt - 5t^2\)
\(v(t) = s'(t) = b- 10t\)
v(t) = 0
b - 10t = 0
b = 10t
Since t = 2.5,
b= 10x2.5 = 25
An object is moving along a line such that its displacement, s at any point in time t is given by
\(s(t) = 2t^3-11t^2+16t-7\)
At what time does the object first change direction.
t =
The object changes direction when s(t) is a maximum or minimum
\(s(t) = 2t^3-11t^2+16t-7\)
\(s'(t) = 6t^2-22t+16\)
Solve \(s'(t) = 0\)
\(6t^2-22t+16=0\\ 3t^2-11t+8=0\)
(3t - 8)(t - 1) = 0
\(t = \frac{8}{3}\quad,\quad t=1\)
Let's check that s(1) is a maximum or minimum
\(s'(t) = 6t^2-22t+16\)
\(s''(t) = 12t-22\)
\(s''(1) = 12-22<0\) Maximum
The object first changes direction when t = 1
An object is moving along a line such that its displacement, s at any point in time t is given by
\(s(t) = sint+cost\)
What is the minimum velocity of the object?
The minimum velocity occurs when acceleration = 0
\(s(t) = sint+cost\)
\(v(t)=s'(t) = cost-sint\)
\(a(t)=v'(t)=s''(t) = -sint-cost\)
a(t) = 0
-sint - cost = 0
sint = -cost
\(\frac{sint}{cost}=-1\)
tant = -1
\(t = \frac{3\pi}{4}\)
Minimum velocity \(v( \frac{3\pi}{4}) = cos( \frac{3\pi}{4})-sin( \frac{3\pi}{4})\)
\(v( \frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} -\frac{1}{\sqrt{2}}\)
\(=-\frac{2}{\sqrt{2}}\\=-\sqrt{2}\)
text
You can get further practice by trying a dynamic quiz.
If you refresh this page, you will get a new set of quizzes
\(\frac{d}{dx}\frac{2}{(3x-5)^{4}} \)
It might help to write question as \(\frac{d}{dx}{2}{(3x-5)^{-4}} \)
If \(f(x)\ =\ \frac { 1 }{ x } -\frac { 3 }{ { x }^{ 2 } } \) then \(f'(x)=\frac { -1 }{ { x }^{ a } } +\frac { b }{ { x }^{ c } } \)
a =
b =
c =
\(f(x)=\frac { 1 }{ x } -\frac { 3 }{ { x }^{ 2 } } ={ x }^{ -1 }-3{ x }^{ -2 }\\ \Rightarrow f'(x)=-1{ x }^{ -2 }-3{ \cdot (-2)x }^{ -3 }=\frac { -1 }{ { x }^{ 2 } } +\frac { 6 }{ { x }^{ 3 } } \)
The following diagram show a graph of f' , the derivative of f
How many stationary points does f have?
Stationary points when \(f' (x) = 0\)
Differentiate sin2x
You should try to remember that \(\frac { d }{ dx } (sinax)=acosax\)
Work out \(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln 2x)\)
Using the laws of logarithms, we know that
\(\large\ln 2x=\ln 2+\ln x\)
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln 2+\ln x)=0+\frac{1}{x}\)
Differentiate \(x^{2}lnx\)
We need to use the product rule
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
u = x² | v = lnx |
\(\frac { du }{ dx } \) = 2x | \(\frac { dv }{ dx } ={1 \over x}\) |
\( \frac{dy}{dx}=x^2 \times{1 \over x} \ +lnx\times 2x\)
= x + 2xlnx
Differentiate \(\frac{x^2+1}{3x-2}\)
We need to use the quotient rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = x² + 1 | v = 3x - 2 |
\(\frac{du}{dx}=2x\) | \(\frac{dv}{dx}=3\) |
\(\frac { dy }{ dx } =\frac { (3x-2)\cdot (2x) -(x^2+1)\cdot 3}{ (3x-2)^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { 6x^2-4x -3x^2-3}{ (3x-2)^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { 3x^2-4x -3}{ (3x-2)^{ 2 } } \)
Differentiate \(\frac { ln3x }{ { e }^{ 2x } } \)
We need to use the Quotient Rule
\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)
u = ln3x | v = \(e^{2x}\) |
\(\frac { du }{ dx } =\frac{1}{x}\) | \(\frac { dv }{ dx } =2e^{2x}\) |
\(\frac { dy }{ dx } =\frac { e^{ 2x }\cdot \frac { 1 }{ x } -2e^{ 2x }\cdot ln3x }{ ({ e }^{ 2x })^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { e^{ 2x }(\frac { 1 }{ x } -2ln3x) }{ ({ e }^{ 2x })^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { \frac { 1 }{ x } -2ln3x }{ { e }^{ 2x } } \)
\(\frac { dy }{ dx } =\frac { \frac { 1 }{ x } -\frac { 2xln3x }{ x } }{ { e }^{ 2x } } \)
\(\frac { dy }{ dx } =\frac { 1-2xln3x }{ { xe }^{ 2x } } \)
The function f(x) = tanx - 4x, \(0 <x<\frac { \pi }{ 2 } \) has a minimum value at \(x = \frac { \pi }{ a} \)
Work out the value of a
a =
Differentiate the function f(x) = tanx - 4x
\(f'(x)=sec^2x-4\)
Minimum value occurs when \(f'(x)=0\)
sec²x - 4 = 0
sec²x = 4
cos²x = \(\frac{1}{4}\)
cosx = \(\pm \frac { 1 }{ 2 } \)
\(Arccos(\frac{1}{2})=\frac{\pi}{3}\)
There is only one correct value in the interval \(0 <x<\frac { \pi }{ 2 } \)
Hence \(x=\frac{\pi}{3}\)
The velocity of a particle moving in a straight line is given by
\(v(t) = 10e^{-0.5t}\)
What is the initial acceleration of the particle?
Initial acceleration =
Acceleration is the rate of change of velocity. We need to differentiate velocity
\(v(t) = 10e^{-0.5t}\)
\(a(t) = v'(t) = 10(-0.5)e^{-0.5t}\)
\(a(t) = -5e^{-0.5t}\)
Initial acceleration is a(0)
\(a(0) = -5e^{-0.5(0)}\)
\(a(0) = -5e^{0}=-5\)
The following is true about a function f
- \(a<b<c\)
- f(x) is increasing on the interval [a,c]
- \(f'(x)<0\) for \(x<a\)
- \(f''(b)=0\)
- \(f'(c)=0\)
Which of the following is TRUE?
Find the equation of the normal to the curve \(y=x^3lnx\) at (1,0)
We need to use the product rule to differentiate \(y=x^3lnx\)
\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)
u = \(x^3\) | v = lnx |
\(\frac{du}{dx}=3x^2\) | \(\frac{dv}{dx}=\frac{1}{x}\) |
\(\frac{dy}{dx}=x^3\cdot \frac{1}{x}+lnx\cdot3x^2\)
\(\frac{dy}{dx}=x^2+3x^2lnx\)
\(\frac{dy}{dx}=x^2(1+lnx)\)
When x = 1, \(\frac{dy}{dx}=1(1+ln1)=1(1+0)=1\)
Gradient of tangent = 1
Gradient of normal = -1
Find the equation of the straight line with gradient = -1 which passes through the point (1,0)
\(\frac{y-0}{x-1}=-1\)
y = -x + 1
The following is true about a function f
- \(a<>
<><><> <><> - f(x) is increasing on the interval [a,c]
- \(f'(x)<0\) for \(x<>
- \(f''(b)=0\)
- \(f'(c)=0\)
Which of the following is TRUE?
Use your graphic display calculator to solve this question
Let \(\large f(x)=sin(e^x) \ , \ 0\le x \le 2\)
The curve of f is concave up on the interval a < x < b
Find the values of a and b
Give your answer to 3 s.f.
a =
b =
The curve of a function is concave up when \(\large f''(x)>0\)
This will occur between two points of inflexion
Differentiate \(\large f(x)=sin(e^x) \) using the Chain Rule
\(\large f'(x)=e^xcos(e^x) \)
We can find the points of inflexion for \(f(x)\) by finding the stationary points for \(f'(x)\)
Use your graphic display calculator to solve this question
Let \(\large f(x)=2x \ln x \ ,\ x>0\)
Points P(1,0) and Q are on the curve of f.
The tangent to the curve of f at P is perpendicular to the tangent to the curve at Q.
Find the x coordinates of Q.
Give your answer to 3 s.f.
a =
You do not need to plot a graph of the curve and the tangents, but it may help you visualise the problem.
Plot the graph of the function and the tangent to the curve at the point x = 1
To get a true picture of the graph, you may want to plot the graph so that the axes are 1:1 (so that a right angle actually looks like a right angle).
You will notice that the gradient of this tangent = 2
You need to find the x coordinates of the graph for which the curve has a gradient = \(-\frac{1}{2}\)
Differentiate the function using the product rule
\(\large f'(x)=2 \ln x+2 \)
Solve \(\large 2 \ln x+2 =-\frac{1}{2}\)
You can use the equation solver or a graphical approach.
Here is what the two tangents look like
How much of HL Mixed Differentiation have you understood?
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