Differentiate with respect to x : 2 3 x + 1

Given, 2 3 x + 1 Let y = 2 3 x + 1 . . . i Now, let u = 3 x + 1 . . . . i i ⇒ y = 2 u . . . . i i i [From i ] From equation i i , we have d u d x = d d x 3 x + 1 = d d x 3 x + d d x 1 x 0 = 3 d d x x + d d x x 0 = 3 × 1 x 1 - 1 + 0 x 0 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = 3 . . . . i v From equation i i i , d y d u = d d u 2 u = 2 d d u u 1 2 = 2 × 1 2 u 1 2 - 1 = u - 1 2 = 1 u 1 2 = 1 u . . . . v We know that d y d x = d y d u × d u d x = 1 u × 3 [using i v , and v ] = 3 u Hence, d d x 2 3 x + 1 = 3 3 x + 1

Differentiate with respect to x : 1 2 x - 5

Given, 1 2 x - 5 Let y = 2 x - 5 - 1 2 . . . i Now, let u = 2 x - 5 . . . . i i ⇒ y = u - 1 2 . . . . i i i [From i ] From equation i i , we have d u d x = d d x 2 x - 5 = d d x 2 x - d d x 5 x 0 = 2 d d x x - 5 d d x x 0 = 2 × 1 x 1 - 1 - 5 × 0 x 0 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = 2 . . . . i v From equation i i i , d y d u = d d u u - 1 2 = - 1 2 u - 1 2 - 1 = - 1 2 u - 3 2 = - 1 2 u 3 . . . . v We know that d y d x = d y d u × d u d x = - 1 2 u 3 × 2 [using i v , and v ] = - 1 u 3 Hence, d d x 1 2 x - 5 = - 1 2 x - 5 3

Differentiate with respect to x : 6 2 - 3 x 3

Given, 6 2 - 3 x 3 Let y = 6 2 - 3 x 3 . . . i Now, let u = 2 - 3 x . . . . i i ⇒ y = 6 u 3 . . . . i i i [From i ] From equation i i , we have d u d x = d d x 2 - 3 x = d d x 2 x 0 - d d x 3 x = 2 d d x x 0 - 3 d d x x = 2 × 0 x 0 - 1 - 3 × 1 x 1 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = - 3 . . . . i v From equation i i i , d y d u = d d u 6 u 3 = 6 d d u 1 u 3 = 6 d d u u - 1 3 = 6 × - 1 3 u - 1 3 - 1 = - 2 u - 4 3 . . . . v We know that d y d x = d y d u × d u d x = - 2 u - 4 3 × - 3 [using i v , and v ] = 6 u - 4 3 = 6 × 1 u 4 3 = 6 u 4 3 Hence, d d x 6 2 - 3 x 3 = 6 2 - 3 x 4 3

Find the gradient of the curve y = x - 3 x + 2 at the points where the curve crosses the x -axis.

Given curve is y = x - 3 x + 2 We know that the gradient of a curve y = f x at point x 1 , y 1 is given by d y d x x 1 , y 1 . So, let us first find d y d x . Consider y = x - 3 x + 2 ⇒ d y d x = d d x x - 3 x + 2 = d d x x - d d x 3 x + 2 = 1 - 3 d d x x + 2 - 1 = 1 - 3 × - 1 x + 2 - 2 ∵ d d x x n = n x n - 1 , n ∈ R = 1 + 3 x + 2 2 . . . . i Now, it is given that the curve crosses the x -axis. So, we find the point of intersection of the given curve with x -axis by substituting y = 0 in the given equation of the curve Thus, y = x - 3 x + 2 ⇒ 0 = x - 3 x + 2 ⇒ x 2 + 2 x - 3 = 0 ⇒ x 2 + 3 x - x - 3 = 0 ⇒ x x + 3 - 1 x + 3 = 0 ⇒ x - 1 x + 3 = 0 ⇒ x = 1 or - 3 Thus, the curve crosses the x -axis at points - 3 , 0 , and 1 , 0 . Now, d y d x = 1 + 3 x + 2 2 d y d x - 3 , 0 = 1 + 3 - 3 + 2 2 = 1 + 3 = 4 And, d y d x 1 , 0 = 1 + 3 1 + 2 2 = 1 + 1 3 = 4 3 Hence, gradient of the given curve at the points where the curve crosses the x -axis are 4 , and 4 3 .

Find the coordinates of the point on the curve y = x 2 - 10 x + 26 where the gradient is 0 .

Given equation of the curve is y = x 2 - 10 x + 26 . . . . . . i We need to find the gradient of this curve i.e., we need to find d y d x . Let u = x 2 - 10 x + 26 ⇒ y = u [Using equation i ] Now, d u d x = d d x x 2 - 10 x + 26 = d d x x 2 - 10 d d x x + d d x 26 x 0 = 2 x 2 - 1 - 10 × 1 x 1 - 1 + 26 × 0 x 0 - 1 = 2 x - 10 . . . . i i And, d y d u = d d u u 1 2 = 1 2 u 1 2 - 1 = 1 2 u - 1 2 = 1 2 u ⇒ d y d u = 1 2 x 2 - 10 x + 26 . . . . . i i i Thus, d y d x = d y d u × d u d x = 1 2 x 2 - 10 x + 26 × 2 x - 10 [Using i i , and i i i ] = x - 5 x 2 - 10 x + 26 . . . . . i v Now, we need to find the coordinates of the point on the curve where d y d x is 0 . ⇒ x - 5 = 0 [From i v ] ⇒ x = 5 Now, substituting the value of x in the given equation of the curve, we have y = 5 2 - 10 × 5 + 26 = 25 - 50 + 26 = 1 = 1 Hence, the required coordinates of the point on the curve where the gradient is zero are 5 , 1

The curve y = a b x - 1 passes through the point 2 , 1 and has gradient - 3 5 at this point. Find the value of a and the value of b .

Given that the curve y = a b x - 1 passes through the point 2 , 1 , and has gradient ' - 3 5 ' Since the point 2 , 1 lies on the curve y = a b x - 1 . So, the coordinates of this point satisfies the equation of this curve. So, substituting x = 2 , and y = 1 in the given equation of the curve, we have 1 = a 2 b - 1 ⇒ a = 2 b - 1 ⇒ - a + 2 b = 1 . . . . i Now, let us find : d y d x d y d x = d d x a b x - 1 = a d d x 1 b x - 1 = a d d x b x - 1 - 1 = a × - 1 b x - 1 - 2 × d d x b x - 1 = - a b x - 1 - 2 × b [ ∵ d d x x n = n x n - 1 , where n is any real number, and d constant d x = 0 ] = - a b b x - 1 2 . . . . i i It is given that d y d x = - 3 5 at 2 , 1 ⇒ - 3 5 = - a b 2 b - 1 2 [From i i ] ⇒ 3 2 b - 1 2 = 5 a b ⇒ 3 4 b 2 - 4 b + 1 = 5 a b ⇒ 12 b 2 - 12 b + 3 = 5 a b . . . . i i i From equation i , we have - a + 2 b = 1 ⇒ - a = 1 - 2 b ⇒ a = 2 b - 1 Substituting the value of a in equation i i i , we have 12 b 2 - 12 b + 3 = 5 2 b - 1 b ⇒ 12 b 2 - 12 b + 3 = 10 b 2 - 5 b ⇒ 2 b 2 - 7 b + 3 = 0 ⇒ 2 b 2 - 6 b - b + 3 = 0 ⇒ 2 b b - 3 - 1 b - 3 = 0 ⇒ 2 b - 1 b - 3 = 0 ⇒ b = 1 2 , 3 From equation i - a + 2 × 1 2 = 1 ⇒ a = 0 But we cannot consider a = 0 , and b = 1 2 . Since y = a b x - 1 = 0 when a = 0 but we need to find the values of a , and b for which y = 1 as the curve passes through the point 2 , 1 . Also, - a + 2 × 3 = 1 ⇒ - a = 1 - 6 = - 5 ⇒ a = 5 Hence, the required values are a = 5 , and b = 3

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B

Sue Pemberton Solutions for Chapter: Differentiation, Exercise 3: EXERCISE 7B

Author:Sue Pemberton

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton Solutions for Chapter: Differentiation, Exercise 3: EXERCISE 7B

Attempt the practice questions on Chapter 7: Differentiation, Exercise 3: EXERCISE 7B with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton Solutions for Chapter: Differentiation, Exercise 3: EXERCISE 7B with Hints & Solutions

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B>Q 3

Differentiate with respect to $x$ :

$2 \sqrt{3 x + 1}$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B>Q 3

Differentiate with respect to $x$ :

$\frac{1}{\sqrt{2 x - 5}}$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B>Q 3

Differentiate with respect to $x$ :

$(i i i)$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B>Q 4

Find the gradient of the curve $y = (2 x - 3)^{5}$ at the point $(2, 1)$ .

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B>Q 5

Find the gradient of the curve $= \frac{1}{u^{\frac{1}{2}}}$ at the point where the curve crosses the $y$ -axis.

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B>Q 6

Find the gradient of the curve $y = x - \frac{3}{x + 2}$ at the points where the curve crosses the $x$ -axis.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B>Q 7

Find the coordinates of the point on the curve $y = \sqrt{(x^{2} - 10 x + 26)}$ where the gradient is $0$ .

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 7 - Differentiation>EXERCISE 7B>Q 8

The curve $y = \frac{a}{b x - 1}$ passes through the point $(2, 1)$ and has gradient $- \frac{3}{5}$ at this point. Find the value of $a$ and the value of $b$ .

Learn and Prepare for any exam you want !

|

|

|

|

|

|

|

|