topic badge
Standard Level

11.10 Quotient Rule

Worksheet
Quotient rule
1

To differentiate y = \dfrac{3 x + 4}{4 x - 3} using the quotient rule, let u = 3 x + 4 and v = 4 x - 3, then:

a

Find u'.

b

Find v'.

c

Hence, find y'.

d

Is it possible for the derivative of this function to be zero?

2

To differentiate y = \dfrac{5 x}{8 x - 9} using the quotient rule, let u = 5 x and v = 8 x - 9, then:

a

Find u'.

b

Find v'.

c

Hence, find y'.

d

Is it possible for the derivative of this function to be zero?

3

To differentiate y = \dfrac{9 x^{2}}{2 x + 8} using the quotient rule, let u = 9 x^{2} and v = 2 x + 8, then:

a

Find u'.

b

Find v'.

c

Hence, find y'.

d

Is it possible for the derivative of this function to be zero?

4

To differentiate y = \dfrac{4 x^{2} + 3}{2 x^{2} + 5} using the quotient rule, let u = 4 x^{2} + 3 and v = 2 x^{2} + 5, then:

a

Find u'.

b

Find v'.

c

Hence, find y'.

d

Is it possible for the derivative of this function to be zero?

5

Differentiate each of the following:

a

y = \dfrac{4 x + 3}{4 x - 3}

b

y = \dfrac{4 x - 3}{3 x - 4}

c

y = \dfrac{2 x}{4 x^{2} - 3}

d

y = \dfrac{5 x + 3}{4 x^2 - 1}

e

y = \dfrac{2 - x}{7 x +3}

f

y = \dfrac{x^2}{4 x^{3} +1}

g

y = \dfrac{3 x + 2}{6 + 5 x^{2}}

h

y = \dfrac{\sqrt{x} + 2}{9 \sqrt{x} + 8}

6

Find the gradient function of y = \dfrac{x^{2} - 5 x - 2}{x + 2}.

7

Consider the function y = \dfrac{3}{x}.

a

By first rewriting it in negative index form, differentiate y.

b

By using the substitutions u = 3 and v = x, differentiate y using the quotient rule.

c

Find the value of x for which the gradient is undefined.

8

Consider the function y = \dfrac{1}{8 x + 3}.

a

Differentiate the function using the chain rule.

b

Differentiate the function using the quotient rule.

9

Consider the function y = \dfrac{3 - 4 x}{3 x - 4}.

a

Differentiate y.

b

Is it possible for the derivative to be zero?

c

Find the value of x that will make the derivative undefined.

10

For each of the following functions:

i

Differentiate y.

ii

Is it possible for the derivative to be zero?

iii

Find the value of x that will make the derivative undefined.

iv

State the region(s) of the domain for which the function is increasing.

v

State the region(s) of the domain for which the function is decreasing.

a

y = \dfrac{x + 6}{x - 6}

b

y = \dfrac{5 x}{3 x - 4}

c

y = \dfrac{4 x}{5-2x}

d

y = \dfrac{2 x + 3}{2 x - 3}

11

Differentiate each of the following:

a

y = \dfrac{\left( 4 t^{2} + 3\right)^{3}}{\left(5 + 2 t\right)^{5}}

b

y = \sqrt{\dfrac{2 + 7 x}{2 - 7 x}}

Gradients and tangents
12

Find f' \left( 4 \right) for f \left( x \right) = \dfrac{5 x}{16 + x^{2}}.

13

Find f' \left( 1 \right) for f \left( x \right) = \dfrac{8}{2 + 2 x^{2}}.

14

Find the value of f' \left( 0 \right) if f \left( x \right) = \dfrac{x}{\sqrt{16 - x^{2}}}.

15

Consider the function f \left( x \right) = \dfrac{4 x^{9}}{\left(x + 2\right)^{4}}.

a

Find f' \left( 2 \right).

b

Is the function increasing or decreasing at x = 2?

16

Find the gradient of f \left( x \right) = \dfrac{\left(x + 7\right)^{9}}{\left(x + 9\right)^{4}} at x = - 5.

17

Find the gradient of the tangent to the curve y = \dfrac{9 x}{4 x + 1} at the point \left(1, \dfrac{9}{5}\right).

18

Find the equation of the tangent to y = \dfrac{x}{x + 4} at the point \left(8, \dfrac{2}{3}\right).

19

Find the equation of the tangent to y = \dfrac{x^{2} - 1}{x + 3} at the point where x = 4.

20

Consider the function g \left( x \right) defined as g \left( x \right) = \dfrac{f \left( x \right)}{x^{3} + 3}, where f \left( x \right) is a function of x.

Given that f \left( 2 \right) = 5 and f' \left( 2 \right) = 6, determine the value of g' \left( 2 \right).

21

Find the values of x such that the gradient of the tangent to the curve y = \dfrac{6 x - 1}{3 x - 1} is - 3.

22

Differentiate y = \dfrac{x^{2}}{x + 3} and find the value of a if y' = 0 at x = a.

23

Differentiate y = \dfrac{x^{2} + k}{x^{2} - k} and find the possible values of k given that y' = 1 at x = - 3.

Sign up to access Worksheet
Get full access to our content with a Mathspace account

What is Mathspace

About Mathspace