HL Mixed Differentiation

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.


Key Concepts

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

Test Yourself

Here's a quiz that practises differentiating xn


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Here's a quiz that practises differentiating functions in the form f(x) + g(x)


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Here's a quiz that practises differentiating functions in the form sinx, cosx, ex and lnx


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Here's a quiz that mainly practises the Chain Rule


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Here's a quiz that mainly practises the Chain Rule and the Product and Quotient Rule


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Here's a quiz that mainly practises the Product and Quotient Rule


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 Here's a quiz that mainly practises the Related Rates of Change


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Here's a quiz that practises differentiating using a graphical display calculator


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Here's a quiz that practises Graphs and the Gradient Function


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Here's a quiz that practises Stationary Points and Equations of Tangents and Normals


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Here's a quiz that practises Kinematics


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Test yourself further

You can get further practice by trying a dynamic quiz.

If you refresh this page, you will get a new set of quizzes

 Easy

1

\(\frac{d}{dx}\frac{2}{(3x-5)^{4}} \)

It might help to write question as \(\frac{d}{dx}{2}{(3x-5)^{-4}} \)

2

If \(f(x)\ =\ \frac { 1 }{ x } -\frac { 3 }{ { x }^{ 2 } } \) then \(f'(x)=\frac { -1 }{ { x }^{ a } } +\frac { b }{ { x }^{ c } } \)

a =

b =

c =

\(f(x)=\frac { 1 }{ x } -\frac { 3 }{ { x }^{ 2 } } ={ x }^{ -1 }-3{ x }^{ -2 }\\ \Rightarrow f'(x)=-1{ x }^{ -2 }-3{ \cdot (-2)x }^{ -3 }=\frac { -1 }{ { x }^{ 2 } } +\frac { 6 }{ { x }^{ 3 } } \)

3

The following diagram show a graph of f' , the derivative of f

How many stationary points does f have?

Stationary points when \(f' (x) = 0\)

4

Differentiate sin2x

You should try to remember that \(\frac { d }{ dx } (sinax)=acosax\)

5

Work out \(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln 2x)\)

Using the laws of logarithms, we know that

\(\large\ln 2x=\ln 2+\ln x\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln 2+\ln x)=0+\frac{1}{x}\)


Medium

1

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 = x²v = 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

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 + 1v = 3x - 2
\(\frac{du}{dx}=2x\)\(\frac{dv}{dx}=3\)

\(\frac { dy }{ dx } =\frac { (3x-2)\cdot (2x) -(x^2+1)\cdot 3}{ (3x-2)^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { 6x^2-4x -3x^2-3}{ (3x-2)^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { 3x^2-4x -3}{ (3x-2)^{ 2 } } \)

3

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 = ln3xv = \(e^{2x}\)
\(\frac { du }{ dx } =\frac{1}{x}\)

\(\frac { dv }{ dx } =2e^{2x}\)

\(\frac { dy }{ dx } =\frac { e^{ 2x }\cdot \frac { 1 }{ x } -2e^{ 2x }\cdot ln3x }{ ({ e }^{ 2x })^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { e^{ 2x }(\frac { 1 }{ x } -2ln3x) }{ ({ e }^{ 2x })^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { \frac { 1 }{ x } -2ln3x }{ { e }^{ 2x } } \)

\(\frac { dy }{ dx } =\frac { \frac { 1 }{ x } -\frac { 2xln3x }{ x } }{ { e }^{ 2x } } \)

\(\frac { dy }{ dx } =\frac { 1-2xln3x }{ { xe }^{ 2x } } \)

4

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 \(x=\frac{\pi}{3}\)

5

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)

\(a(0) = -5e^{-0.5(0)}\)

\(a(0) = -5e^{0}=-5\)


 Hard

1

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?

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

3

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?

4

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 \(f'(x)\)

5

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


MY PROGRESS

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