Answer
$f'(x)=\frac{-9 (5x^4-1)}{(x^5-x+1)^{10}} $
Work Step by Step
Given that $f(x)=(x^5-x+1)^{-9} $
We can first write it as $ f ( x ) =(x^5-x+1)^{-9} $ so that it is easier to deal with.
Next we write this as$ f(x)=(g(x))^{-9}$, where $f(x) $ is the outside function and $ g(x)=x^5-x+1$ is the inside function.
To find the answer we need to differentiate the outside function $f(x)$ and times it by the derivative of the inside function $ g ( x ) $.
$f(x)=(x^5-x+1)^{-9} $
$ f (x)=(g (x))^{-9}$
$f'(x)=-9(g ( x ))^{-9-1}×(g' ( x )) $
$f'(x)=-9(g ( x ))^{-10}×(5x^{5-1}-1$
$f'(x)=-9(x^5-x+1)^{-10}×(5x^{4}-1) $
$f'(x)=-9(5x^{4}-1)(x^5-x+1)^{-10}$
$f'(x)=\frac{-9 (5x^4-1)}{(x^5-x+1)^{10}}$
