Sec 4 A-Math Differentiation of trigonometric, exponential and logarithmic functions worksheet with answers

A free Sec 4 A-Math worksheet on differentiation of trigonometric, exponential and logarithmic functions: 7 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 4 A-Math worksheet: Differentiation of trigonometric, exponential and logarithmic functions

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

Given that

y = x\cos x

, where

x

is in radians, find

\dfrac{dy}{dx}

Answer[3]

Given that

y = 3\ln(x^{2} + 3)

, find

\dfrac{dy}{dx}

Answer[3]

(a)Given that

y = \ln(3x - 1)

, find

\dfrac{dy}{dx}

Answer (a)[3]

(b)Given that

y = \ln\left(\dfrac{2x - 3}{x + 1}\right)

, find

\dfrac{dy}{dx}

Answer (b)[1]

(a)Given that

y = e^{-x - 3}

, find

\dfrac{dy}{dx}

Answer (a)[3]

(b)Given that

y = 3e^{-x^{2} - 2x}

, find

\dfrac{dy}{dx}

Answer (b)[3]

Section B

A curve has equation

y = 2x + \dfrac{50}{x}

, for

x > 0

(a)Find

\dfrac{d^{2}y}{dx^{2}}

Answer (a)[2]

(b)Without finding its coordinates, state the nature of any stationary point of the curve, explaining your answer.

Answer (b)[1]

(c)Find the

x

-coordinate of the stationary point.

Answer (c)[2]

(d)Find the range of values of

x

for which

y

is an increasing function.

Answer (d)[1]

A curve has equation

y = \dfrac{1 - \cos 2x}{\sin 2x}

, for

0 < x < \dfrac{\pi}{2}

(a)Show that

\dfrac{1 - \cos 2x}{\sin 2x} = \tan x

[3]

(b)Hence find

\dfrac{dy}{dx}

Answer (b)[1]

(c)Find the equation of the tangent to the curve at the point where

x = \dfrac{\pi}{4}

, leaving your answer in exact form.

Answer (c)[3]

(d)Find the value of

x

for which the gradient of the curve is

\dfrac{4}{3}

Answer (d)[2]

The equation of a curve is

y = (x - 2)e^{-2x}

(a)Find

\dfrac{dy}{dx}

Answer (a)[2]

(b)Find the exact coordinates of the stationary point of the curve.

Answer (b)[3]

(c)Determine the nature of the stationary point.

Answer (c)[2]

(d)Find the range of values of

x

for which

y

is increasing.

Answer (d)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= -x\sin x + \cos x

[3]

Use the product rule.

u = x

v = \cos x

\dfrac{du}{dx} = 1

\dfrac{dv}{dx} = -\sin x

\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}

\dfrac{dy}{dx} = x(-\sin x) + \cos x \cdot 1

= -x\sin x + \cos x

Question 2

Answer:

= \dfrac{6x}{x^{2} + 3}

[3]

u = x^{2} + 3

\dfrac{du}{dx} = 2x

\dfrac{dy}{du} = \dfrac{3}{u}

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

\dfrac{dy}{dx} = \dfrac{3}{x^{2} + 3} \times 2x

= \dfrac{6x}{x^{2} + 3}

Question 3

(a)Answer:

= \dfrac{3}{3x - 1}

[3]

u = 3x - 1

\dfrac{du}{dx} = 3

\dfrac{dy}{du} = \dfrac{1}{u}

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

\dfrac{dy}{dx} = \dfrac{1}{3x - 1} \times 3

= \dfrac{3}{3x - 1}

(b)Answer:

\dfrac{dy}{dx} = \dfrac{2}{2x - 3} - \dfrac{1}{x + 1}

[1]

y = \ln(2x - 3) - \ln(x + 1)

\dfrac{dy}{dx} = \dfrac{2}{2x - 3} - \dfrac{1}{x + 1}

Question 4

(a)Answer:

= -e^{-x - 3}

[3]

u = -x - 3

\dfrac{du}{dx} = -1

\dfrac{dy}{du} = e^{u}

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

\dfrac{dy}{dx} = e^{-x - 3} \times (-1)

= -e^{-x - 3}

(b)Answer:

= 3(-2x - 2)e^{-x^{2} - 2x}

[3]

u = -x^{2} - 2x

\dfrac{du}{dx} = -2x - 2

\dfrac{dy}{du} = 3e^{u}

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

\dfrac{dy}{dx} = 3e^{-x^{2} - 2x} \times (-2x - 2)

= 3(-2x - 2)e^{-x^{2} - 2x}

Question 5

(a)Answer:

\dfrac{d^{2}y}{dx^{2}} = \dfrac{100}{x^{3}}

[2]

\dfrac{dy}{dx} = 2 - 50x^{-2}

\dfrac{d^{2}y}{dx^{2}} = 100x^{-3}

= \dfrac{100}{x^{3}}

(b)Answer: Minimum[1]

For

x > 0

x^{3} > 0

, so

\dfrac{d^{2}y}{dx^{2}} = \dfrac{100}{x^{3}} > 0

.So any stationary point is a minimum point.

(c)Answer:

x = 5

[2]

2 - \dfrac{50}{x^{2}} = 0

x^{2} = 25

x = 5

(as

x > 0

)

(d)Answer:

x > 5

[1]

\dfrac{dy}{dx} = 2 - \dfrac{50}{x^{2}} > 0

when

x^{2} > 25

x > 5

Question 6

(a)[3]

\dfrac{1 - \cos 2x}{\sin 2x} = \dfrac{1 - (1 - 2\sin^{2} x)}{2\sin x\cos x}

= \dfrac{2\sin^{2} x}{2\sin x\cos x}

= \dfrac{\sin x}{\cos x}

= \tan x

(shown)

(b)Answer:

\dfrac{dy}{dx} = \sec^{2} x

[1]

y = \tan x

\dfrac{dy}{dx} = \sec^{2} x

(c)Answer:

y - 1 = 2\left(x - \dfrac{\pi}{4}\right)

[3]

When

x = \dfrac{\pi}{4}

y = \tan \dfrac{\pi}{4}

y = 1

\dfrac{dy}{dx} = \sec^{2} \dfrac{\pi}{4}

\phantom{\dfrac{dy}{dx}} = 2

y - 1 = 2\left(x - \dfrac{\pi}{4}\right)

(d)Answer:

x = \dfrac{\pi}{6}

[2]

\sec^{2} x = \dfrac{4}{3}

\cos^{2} x = \dfrac{3}{4}

For

0 < x < \dfrac{\pi}{2}

x = \dfrac{\pi}{6}

Question 7

(a)Answer:

\dfrac{dy}{dx} = e^{-2x}(-2x + 5)

[2]

\dfrac{dy}{dx} = (1)e^{-2x} + (x - 2)(-2e^{-2x})

= e^{-2x}(-2x + 5)

(b)Answer:

\left(\dfrac{5}{2}, \dfrac{1}{2}e^{-5}\right)

[3]

e^{-2x}(-2x + 5) = 0

e^{-2x} > 0

for all

x

, so

-2x + 5 = 0

x = \dfrac{5}{2}

y = \left(\dfrac{5}{2} - 2\right)e^{-5}

= \dfrac{1}{2}e^{-5}

(c)Answer: Maximum point[2]

\dfrac{d^{2}y}{dx^{2}} = -2e^{-2x}(-2x + 5) + e^{-2x}(-2)

= e^{-2x}(4x - 12)

When

x = \dfrac{5}{2}

\dfrac{d^{2}y}{dx^{2}} = -2e^{-5}

This is negative, so the point is a maximum.

(d)Answer:

x < \dfrac{5}{2}

[2]

y

increases where

\dfrac{dy}{dx} > 0

e^{-2x} > 0

, so

-2x + 5 > 0

x < \dfrac{5}{2}

Practice questions by skill

Sec 4 A-Math Differentiation of trigonometric, exponential and logarithmic functions worksheet with answers

In the print window, choose Save as PDF to download it.

Section B

A curve has equation

y = 2x + \dfrac{50}{x}

, for

x > 0

(a)Find

\dfrac{d^{2}y}{dx^{2}}

Answer (a)[2]

(b)Without finding its coordinates, state the nature of any stationary point of the curve, explaining your answer.

Answer (b)[1]

(c)Find the

x

-coordinate of the stationary point.

Answer (c)[2]

(d)Find the range of values of

x

for which

y

is an increasing function.

Answer (d)[1]

Answers and worked solutions

Question 1

Answer:

= -x\sin x + \cos x

[3]

Use the product rule.

u = x

v = \cos x

\dfrac{du}{dx} = 1

\dfrac{dv}{dx} = -\sin x

\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}

\dfrac{dy}{dx} = x(-\sin x) + \cos x \cdot 1

= -x\sin x + \cos x

Question 2

Answer:

= \dfrac{6x}{x^{2} + 3}

[3]

u = x^{2} + 3

\dfrac{du}{dx} = 2x

\dfrac{dy}{du} = \dfrac{3}{u}

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

\dfrac{dy}{dx} = \dfrac{3}{x^{2} + 3} \times 2x

= \dfrac{6x}{x^{2} + 3}

Question 3

(a)Answer:

= \dfrac{3}{3x - 1}

[3]

u = 3x - 1

\dfrac{du}{dx} = 3

\dfrac{dy}{du} = \dfrac{1}{u}

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

\dfrac{dy}{dx} = \dfrac{1}{3x - 1} \times 3

= \dfrac{3}{3x - 1}

(b)Answer:

\dfrac{dy}{dx} = \dfrac{2}{2x - 3} - \dfrac{1}{x + 1}

[1]

y = \ln(2x - 3) - \ln(x + 1)

\dfrac{dy}{dx} = \dfrac{2}{2x - 3} - \dfrac{1}{x + 1}

Question 4

(a)Answer:

= -e^{-x - 3}

[3]

u = -x - 3

\dfrac{du}{dx} = -1

\dfrac{dy}{du} = e^{u}

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

\dfrac{dy}{dx} = e^{-x - 3} \times (-1)

= -e^{-x - 3}

(b)Answer:

= 3(-2x - 2)e^{-x^{2} - 2x}

[3]

u = -x^{2} - 2x

\dfrac{du}{dx} = -2x - 2

\dfrac{dy}{du} = 3e^{u}

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

\dfrac{dy}{dx} = 3e^{-x^{2} - 2x} \times (-2x - 2)

= 3(-2x - 2)e^{-x^{2} - 2x}

Question 5

(a)Answer:

\dfrac{d^{2}y}{dx^{2}} = \dfrac{100}{x^{3}}

[2]

\dfrac{dy}{dx} = 2 - 50x^{-2}

\dfrac{d^{2}y}{dx^{2}} = 100x^{-3}

= \dfrac{100}{x^{3}}

(b)Answer: Minimum[1]

For

x > 0

x^{3} > 0

, so

\dfrac{d^{2}y}{dx^{2}} = \dfrac{100}{x^{3}} > 0

.So any stationary point is a minimum point.

(c)Answer:

x = 5

[2]

2 - \dfrac{50}{x^{2}} = 0

x^{2} = 25

x = 5

(as

x > 0

)

(d)Answer:

x > 5

[1]

\dfrac{dy}{dx} = 2 - \dfrac{50}{x^{2}} > 0

when

x^{2} > 25

x > 5

Question 6

(a)[3]

\dfrac{1 - \cos 2x}{\sin 2x} = \dfrac{1 - (1 - 2\sin^{2} x)}{2\sin x\cos x}

= \dfrac{2\sin^{2} x}{2\sin x\cos x}

= \dfrac{\sin x}{\cos x}

= \tan x

(shown)

(b)Answer:

\dfrac{dy}{dx} = \sec^{2} x

[1]

y = \tan x

\dfrac{dy}{dx} = \sec^{2} x

(c)Answer:

y - 1 = 2\left(x - \dfrac{\pi}{4}\right)

[3]

When

x = \dfrac{\pi}{4}

y = \tan \dfrac{\pi}{4}

y = 1

\dfrac{dy}{dx} = \sec^{2} \dfrac{\pi}{4}

\phantom{\dfrac{dy}{dx}} = 2

y - 1 = 2\left(x - \dfrac{\pi}{4}\right)

(d)Answer:

x = \dfrac{\pi}{6}

[2]

\sec^{2} x = \dfrac{4}{3}

\cos^{2} x = \dfrac{3}{4}

For

0 < x < \dfrac{\pi}{2}

x = \dfrac{\pi}{6}

Question 7

(a)Answer:

\dfrac{dy}{dx} = e^{-2x}(-2x + 5)

[2]

\dfrac{dy}{dx} = (1)e^{-2x} + (x - 2)(-2e^{-2x})

= e^{-2x}(-2x + 5)

(b)Answer:

\left(\dfrac{5}{2}, \dfrac{1}{2}e^{-5}\right)

[3]

e^{-2x}(-2x + 5) = 0

e^{-2x} > 0

for all

x

, so

-2x + 5 = 0

x = \dfrac{5}{2}

y = \left(\dfrac{5}{2} - 2\right)e^{-5}

= \dfrac{1}{2}e^{-5}

(c)Answer: Maximum point[2]

\dfrac{d^{2}y}{dx^{2}} = -2e^{-2x}(-2x + 5) + e^{-2x}(-2)

= e^{-2x}(4x - 12)

When

x = \dfrac{5}{2}

\dfrac{d^{2}y}{dx^{2}} = -2e^{-5}

This is negative, so the point is a maximum.

(d)Answer:

x < \dfrac{5}{2}

[2]

y

increases where

\dfrac{dy}{dx} > 0

e^{-2x} > 0

, so

-2x + 5 > 0

x < \dfrac{5}{2}