Answer
f'(x)=sec(x)((2$x^{4}$-4$x^{-1}$)(tan(x)+(8$x^{3}$+4$x^{-2}$))
Work Step by Step
The product rule states that if f(x)=h(x)g(x), then f'(x)=h(x)g'(x)+g(x)h'(x). Since f(x)=(2$x^{4}$-4$x^{-1}$)sec(x), we can set h(x)=(2$x^{4}$-4$x^{-1}$) and g(x)=sec(x), then apply the product rule to find f'(x). Therefore,
f'(x)=(2$x^{4}$-4$x^{-1}$)$\frac{d}{dx}$[sec(x)]+(sec(x))$\frac{d}{dx}$[2$x^{4}$-4$x^{-1}$]
We know that
$\frac{d}{dx}$[2$x^{4}$-4$x^{-1}$]=8$x^{3}$+4$x^{-2}$, using the power rule
$\frac{d}{dx}$[sec(x)]=sec(x)tan(x), which can be derived by using the quotient rule to compute the derivative of the equation g(x)=$\frac{1}{cos(x)}$, which of course is equivalent to sec(x)
Therefore,
f'(x)=(2$x^{4}$-4$x^{-1}$)(sec(x)tan(x))+(sec(x))(8$x^{3}$+4$x^{-2}$)
=sec(x)((2$x^{4}$-4$x^{-1}$)(tan(x)+(8$x^{3}$+4$x^{-2}$))
