How do you differentiate $f(x)=(2x^2-6x+1)^-8$ ?

Question

2616 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-08T19:12:02

Use the chain rule. Please see explanation for details.

Use the chain rule $(df(u(x)))/dx = ((df)/(du))((du)/dx)$

let $u(x) = 2x² - 6x + 1$ , then $f(u) = u^(-8)$ , $(df(u))/(du) = -8u^(-9)$ , and $(du(x))/(dx) = 2x - 6$

Substituting into the chain rule:

$f'(x) = (-8u^(-9))(2x - 6)$

Reverse the substitution for u:

$f'(x) = -8(2x² - 6x + 1)^(-9)(2x - 6)$

Simplify a bit:

$f'(x) = (48 - 16x)/(2x² - 6x + 1)^(9)$

How do you differentiate f(x)=(2x^2-6x+1)^-8f(x)=(2x^2-6x+1)^-8?