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
- 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!
Here's a quiz that practises differentiating functions in the form f(x) + g(x)
START QUIZ!
\(\large\frac{\mathrm{d}}{\mathrm{d}x}(x^3+x^2)=ax^2+2x^b\)
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!
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!
Here's a quiz that mainly practises the Chain Rule and the Product and Quotient Rule
START QUIZ!
Here's a quiz that mainly practises the Product and Quotient Rule
START QUIZ!
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 practises differentiating using a graphical display calculator
START QUIZ!
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!
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.
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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
- <>
- 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!
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
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EasyThe following diagram show a graph of f' , the derivative of f
How many stationary points does f have?
Stationary points when \(f' (x) = 0\)
f(x) = x3+2x
Find the gradient of the curve y = f(x) at the point (-1,-3)
gradient =
f'(x)=3x²+2
f '(-1) = 3(-1)²+2 = 5
The equation of the tangent to the curve f(x)=x3 at the point (1,1) is given by y = mx -2
Find m
m =
f(x) = x3
f '(x) = 3x²
f '(1) = 3
Gradient of the tangent = 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 \(\int { \frac { 1 }{ 4x } dx= } \frac { 1 }{ 4 } \int { \frac { 1 }{ x } dx= } \frac { 1 }{ 4 } lnx\)
Differentiate sin2x
You should try to remember that \(\frac { d }{ dx } (sinax)=acosax\)
MediumIf \(y=3{ x }^{ 2 }+2\sqrt { x } \) then find \(\frac { dy }{ dx } \) when x = 4
\(y=3{ x }^{ 2 }+2\sqrt { x } =3{ x }^{ 2 }+2{ x }^{ \frac { 1 }{ 2 } }\\ \Rightarrow \frac { dy }{ dx } =\quad 6x+{ x }^{ -\frac { 1 }{ 2 } }=6x+\frac { 1 }{ \sqrt { x } } \\ when\quad x=4\quad ,\quad \frac { dy }{ dx } =24+\frac { 1 }{ \sqrt { 4 } } =24+\frac { 1 }{ 2 } \)
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)
=\(-3x^{2}e^{-x^{3}}\)
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:
\(s(t) = e^{-2t}\)
\(v(t)=s'(t) = -2e^{-2t}\)
\(a(t)=v'(t)=s''(t) = 4e^{-2t}\)
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 \(\large f(x)=2x^2-\sqrt{x}\) , find \(\large f'(0.25)\)
\(\large f'(0.25)\) =
\(\large f(x)=2x^2-x^{\frac{1}{2}}\)
\(\large f'(x)=4x-\frac{1}{2}x^{-\frac{1}{2}}=4x-\frac{1}{2\sqrt{x}}\)
\(\large f'(0.25)=4(0.25)-\frac{1}{2\sqrt{0.25}}=1-\frac{1}{2\times 0.5}=1-1=0\)
HardComplete the following derivative
\(\frac{d}{dx}(\frac {x}{x+3})=\frac { a\cdot 1-x\cdot b }{ { (x+3) }^{ c } } \)
x + 3 1 2 x 3
a =
b =
c =
\(\frac{d}{dx}(\frac {u}{v})=\frac { v\cdot \frac{du}{dx}-u\cdot \frac{dv}{dx} }{ v^2 } \\ \frac{d}{dx}(\frac {x}{x+3})=\frac { (x+3)\cdot 1-x\cdot 1 }{ { (x+3) }^{ c2} } \)
Find the value of a in the following derivative
\(\frac{d}{dx}(\frac {2e^x-1}{2e^x+1})=\frac { ae^x }{ (2e^x+1)^2} \)
a =
\(\frac{d}{dx}(\frac {2e^x-1}{2e^x+1})=\frac { (2e^x+1)2e^x-2e^x(2e^x-1) }{ (2e^x+1)^2} \\\quad \quad \quad \quad=\frac { 4e^{2x}+2e^x-4e^{2x}+2e^x }{ (2e^x+1)^2} \\\quad \quad \quad \quad=\frac { 4e^x }{ (2e^x+1)^2} \\\)
Consider two functions f and g . Given that \(h(x)=\frac{f(x)}{g(x)}\)
The table shows valuse for f , g , h and their derivatives f' , g' and h' for x = 1 and x = 2
Example, f(1) = 3 and g '(2) = -2
Complete the table
| x | 1 | 2 |
|---|---|---|
| f(x) | 3 | |
| f '(x) | -2 | 2 |
| g(x) | -1 | 1 |
| g '(x) | 1 | |
| h(x) | 2 | |
| h '(x) | 6 |
\(h(1)=\frac{f(1)}{g(1)}=\frac{3}{-1}=-3\)
\(h'(1)=\frac{g(1)f'(1)-g'(1)f(1)}{[g(1)]^2}=\frac{-1(-2)-1\cdot3}{[-1]^2}=-1\)
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 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
How much of SL Mixed Differentiation have you understood?
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