Bismillah Adni PreUniversity Cambridge International A Level Programme

Bismillah. . . Adni Pre-University Cambridge International A Level Programme Unit P 1: Pure Mathematics 1 (Paper 1) Topic: Differentiation Chapter 12: Extending Differentiation

s c h • Differentiating composite functions a p t e r

o u t • c o m e s We should be able to: – Differentiate composite functions of the form f(F(x)) – Apply differentiation to rates of change and to related rates of change.

a x + • To differentiate a function like y = (2 x + 1)3, there are 2 methods: b ) 1. Expand the function (can use the binomial n theorem) and then differentiate term by term. : E. g. 12. 1. 1: y = (2 x + 1)3 = 8 x 3 + 12 x 2 + 6 x + 1 M e t h dy/ dx = 24 x 2 + 24 x + 6 = 6(4 x 2 + 4 x + 1) = 6(2 x + 1)2

a x + 2. Use a formula: n, • If a, b and n are constants, and y = (ax + b) b ) then dy/ = n(ax + b)n-1 * a dx n : M e t h or dy/dx = na(ax + b)n-1 E. g. 12. 1. 1: y = (2 x + 1)3 dy/ dx = 3(2 x + 1)3 -1*2 = 6(2 x + 1)2

1 2. dy/ for y = (1 - 3 x)4. • Find 1 dx. 1 :

1 2. dy/ for y = (1 - 3 x)4. • Find 1 dx. • Answer: 1 y = (1 - 3 x)4 : = 81 x 4 - 108 x 3 + 54 x 2 - 12 x + 1 dy/ dx = 324 x 3 - 324 x 2 + 108 x -12 y = (1 - 3 x)4 dy/ dx = 4(1 - 3 x)3 (-3) = -12(1 - 3 x)3 OR

1 2. 1. 2 : • Find dy/dx when y = √(3 x + 2). • Answer: y = √(3 x + 2) = (3 x + 2)½ dy/ = dx ½(3 x + 2)½-1*3 = 3/2 (3 x + 2)-½ =

1 2. 1. 2 : • Find dy/dx when y = 1/(1 - 2 x). • Answer: y = 1/(1 - 2 x) = (1 - 2 x)-1 dy/ dx = (-1)(1 - 2 x)-1 -1*(-2) = 2(1 - 2 x)-2 =

The chain rule • For y = f(u), if u = F(x) and y = f(F(x)): then • We can think about differentiation in terms of rates of change: • dy/ : the rate of change of y with respect to x. dx • dy/ : the rate of change of y with respect to u. du • du/ : the rate of change of u with respect to x. dx

1 2. dy/ when y = (1 + x 2)3. • Find dx 2. 2 • Answer: : • Let u = 1 + x 2, so that y = u 3. • du/ = 2 x and dy/ 2 = 3 u dx du = 3(1 + x 2)2 • dy/ = dy/ * du/ dx du dx = 3(1 + x 2)2 * 2 x = 6 x(1 + x 2)2

r : • y = (1 + x 2)3 = 1 + 3 x 2 + 3 x 4 + x 6 dy/ = 6 x + 12 x 3 + 6 x 5 dx = 6 x(1 + 2 x 2 + x 4) = 6 x(1 + x 2)2 Or: • y = (1 + x 2)3 dy/ = 3(1 + x 2)3 -1*(2 x) dx = 6 x(1 + x 2)2

1 2. ½ with respect to x. • Differentiate y = (2 x + 1) 2. 3 :

1 2. 2. 3 : • • • Differentiate y = (2 x + 1)½ with respect to x. Answer: Let u = 2 x + 1, so that y = u½ du/ = 2 and dy/ ½-1 dx du = ½ u -½ = ½ (2 x + 1)-½ dy/ = dy/ * du/ dx du dx = ½ (2 x + 1)-½ * 2 = (2 x + 1)-½ =

e s o f • We often need to calculate the rate at which one quantity varies with another when one of them is time. c h • If r is some quantity, then the rate of change of r with respect to time t is dr/dt. a n • Problems like this can be answered by using the g chain rule. e

1 2. • Suppose that a spherical balloon is being 3 inflated at a constant rate of 5 m 3/s. At a. particular moment, the radius of the balloon is 4 1 metres. Find how fast the radius of the balloon is increasing at that instant.

Answer: • Let: • V be the volume of the balloon (m 3), • r be its radius (m), • t be the time for which the balloon has been inflating (s). • Given: • d. V/ = 5 m 3/s, dt • r = 4 m. • Find: dr/dt (in m/s).

• Volume of the spherical balloon, V = 4/3 πr 3 • So, d. V/dr = (3*4/3) πr 2 -1 = 4πr 2

The End Alhamdulillah. . .