DIFFERENTIATION TECHNIQUES

with the following standard results: To differentiate various expressions it is necessary to be very familiar with the following standard results: Trigonometric: Exponential: y dy dx y dy dx sin kx k cos kx e kx ke kx cos kx –k sin kx Logarithmic: tan kx k sec2 kx y dy dx sec x sec x tan x cosec x – cosec x cot x f (x) f(x) ln f(x) cot x – cosec2x

It is also necessary to know the following: dy dx = Products: If y = uv, then uv + vu If y = u v , then dy dx = vu – uv v2 Quotients: “Bracket to a power” If y = [ f(x) ] n , then dy dx = n [ f(x) ] n-1 f (x)

dy dx Example 1: Find given that y = x2 sin 4x dy dx = u v dy dx = This is a product, so x2 ( 4 cos 4x ) + sin 4x ( 2x ) u v v u = 2x ( 2x cos 4x + sin 4x ) Example 2: Find dy dx given that y = x e2x u v v u u v dy dx = e2x (1) – x (2e2x) This is a quotient, so (e2x)2 v2 dy dx = 1 – 2x e2x e2x : Divide throughout by

Example 3: Find dy dx given that y = 3 ( 2x + 1 )5. This is a “bracket to a power”. f (x) dy dx = 15( 2x + 1 )4 × 2 = 30 ( 2x + 1 )4 Example 4: Find dy dx given that y = ln ( 1 + 2x2) f (x) dy dx = 4x f(x) 1 + 2x2

If x is given as a function of y, it can sometimes be necessary to use the result: dy d x d y 1 = Example 5: Given that x = tan y, find dy d x in terms of x. x = tan y d x d y = sec2 y dy d x = 1 sec2 y = 1 1 + tan2 y = 1 1 + x2

It is vital to know the following results: Summary of key points: It is vital to know the following results: Trigonometric: Exponential: y dy dx y dy dx sin kx k cos kx e kx ke kx cos kx –k sin kx Logarithmic: tan kx k sec2 kx y dy dx sec x sec x tan x cosec x – cosec x cot x f (x) f(x) ln f(x) cot x – cosec2x

……and the following: dy dx = Products: If y = uv, then uv + vu If y = vu – uv v2 Quotients: “Bracket to a power” If y = [ f(x) ] n , then dy dx = n [ f(x) ] n-1 f (x) This PowerPoint produced by R.Collins ; Updated Apr.2009

This is an example of “implicit differentiation”. dy dx Example 6: Find given that x2 – y3 = 5x + y This is an example of “implicit differentiation”. The method is to differentiate each term with respect to x. However the chain rule is needed for the terms involving y. d dx f(y) = dy f(y) i.e. use the result: In practice, this means differentiating with respect to y and multiplying by dy dx . – 3y2 dy dx dy dx + So: 2x = 5 dy dx = 2x – 5 1 + 3y2 Re-arranging this gives: