Derivatives of Functions of f(x) = u |
|
Considering f(x) = (x + x2)5(1 + x3)2, find f'(x) Let’s re-write the function as follows: y = (x2 + x)5(1 + x3)2 Let’s consider a function k as follows: k = x2 + x k' = 2x + 1 Let’s consider a function h as follows: u = k5 = (x2 + x)5 u' = nkn-1k' = 5(x2 + x)4(2x + 1) = 5(2x + 1)(x2 + x)4 Let’s consider a function g as follows: g = 1 + x3 g’ = 3x2 Let’s consider a function m as follows: v = g2 = (1 + x3)2 v' = ngn-1g' = 2(1 + x3)(3x2) = 6x2(1 + x3) Let’s consider a function f as follows: f = uv To calculate the derivative, we use the formula (uv)' = u'v + uv': f' = u'v + uv' = 5(2x + 1)(x2 + x)4(1 + x3)2 +(x2 + x)56x2(1 + x3) = 5(2x + 1)(x2 + x)4(1 + x3)2 +6x2(1 + x3)(x2 + x)5 = (x2 + x)4(1 + x3)[5(2x + 1)(1 + x3) + 6x2(x2 + x)]
Let’s consider a function k as follows: g = 3x3 + 5 g' = 9x2 Let’s consider a function h as follows: u = g2 = (3x3 + 5)2 u' = ngn-1g' = 3(3x3 + 5)(9x2) = 27x2(3x3 + 5) Let’s consider a function k as follows: k = 4x2 + 2x + 3 k’ = 8x + 2 Let’s consider a function h as follows: v = k4 = (4x2 + 2x + 3)4 v’ = nkn-1k' = 4(4x2 + 2x + 3)3(8x + 2) = 4(8x + 2)(4x2 + 2x + 3)3 Let’s consider a function f as follows: f = u/v = (3x3 + 5)2/(4x2 + 2x + 3)4 f' = (u'v + uv')/v8 Considering f(x) = [3x + (2x + x5)-2]-4, find f'(x) Let’s re-write the function as follows: y = (3x + (x5 + 2x)-2)-4 Let’s consider a function k as follows: k = x5 + 2x k' = 5x4 + 2 Let’s consider a function h as follows: h = k-2 = (x5 + 2x)-2 h' = nkn-1k' = -2( x5 + 2x)-3(5x4 + 2) = -2(5x4 + 2)(x5 + 2x)-3 Let’s consider a function g as follows: g = 3x + h = 3x + (x5 + 2x)-2 g’ = (3x)' + h' = 3 + (-2(5x4 + 2)(x5 + 2x)-3) = -2(5x4 + 2)(x5 + 2x)-3 + 3 Let’s consider a function f as follows: f = g-4 = [3x + (x5 + 2x)-2]-4 f' = ngn-1g' = -4[3x + (x5 + 2x)-2]-5[-2(5x4 + 2)(x5 + 2x)-3 + 3] = -4[-2(5x4 + 2)(x5 + 2x)-3 + 3][3x + (x5 + 2x)-2]-5
|
|
||
Home | Copyright © 2007-2013, FunctionX | |
|