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

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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)#