How do you differentiate f(x)= (4x^5+5)^(1/2)f(x)=(4x5+5)12 using the chain rule?

2 Answers
May 22, 2018

{d[f(x)]}/{dx} = {10x^4}/{sqrt{4x^5 + 5}}d[f(x)]dx=10x44x5+5

Explanation:

f(x) = (4x^5 + 5)^{1/2}f(x)=(4x5+5)12

{d[f(x)]}/{dx} = d/dx[(4x^5 + 5)^{1/2}]d[f(x)]dx=ddx[(4x5+5)12]

let u = 4x^5 + 5u=4x5+5

{d[f(x)]}/{dx} = d/{du}[u^{1/2}]d/dx[4x^5 + 5]d[f(x)]dx=ddu[u12]ddx[4x5+5]

{d[f(x)]}/{dx} = 1/2u^{-1/2} times 20x^4d[f(x)]dx=12u12×20x4

Substitute uu back in

{d[f(x)]}/{dx} = 1/2( 4x^5 + 5)^{-1/2} times 20x^4d[f(x)]dx=12(4x5+5)12×20x4

Simplify

{d[f(x)]}/{dx} = {10x^4}/{sqrt{4x^5 + 5}}d[f(x)]dx=10x44x5+5

May 22, 2018

f'(x)=(10x^4)/(4x^5+5)^(1/2)

Explanation:

"given "f(x)=g(h(x))" then"

f'(x)=g'(h(x))xxh'(x)larrcolor(blue)"chain rule"

rArrf'(x)=1/2(4x^5+5)^(-1/2)xxd/dx(4x^5+5)

"color(white)(rArrf'(x))=1/cancel(2)(4x^5+5)^(-1/2)xxcancel(20)^(10)x^4

color(white)(rArrf'(x))=(10x^4)/(4x^5+5)^(1/2)