How do I solve f'(x) for f(x)=7(x^2)+(2/(x^2))?

I thought it would become... 14x+2x^-2 --> 14+-4x^-3 --> (-4/(x^3))+14, but according to the homework system I use online, this does not seem to be correct. I'm trying to use the power rules, but don't think I'm completely understand them. Thanks for your help!

1 Answer
Feb 24, 2018

f^{'}(x)=14x-\frac{4}{x^3}

Write the answer in this exact format in your homework system.

Explanation:

f^'(x)=\frac{d}{dx}(7x^2+\frac{2}{x^2})

\text{Apply the Sum/Difference of derivative Rule:}

\quad (f\pm g)^'=f^'\pm g^'

=\frac{d}{dx}(7x^2)+\frac{d}{dx}(\frac{2}{x^2})

By taking the constant out (a\cdot f(x))^'=a\cdot f^{'}(x) and applying the power rule \ \ \ \ \frac{d}{dx}(x^a)=a\cdot x^{a-1},\ \ \ \ we get:

\frac{d}{dx}(7x^2)=7\frac{d}{dx}(x^2)=7\cdot 2x^{2-1}=14x

\frac{d}{dx}(\frac{2}{x^2})=2\frac{d}{dx}(x^{-2})=2(-2x^{-2-1})=-\frac{4}{x^3}

So that we get:

=14x-\frac{4}{x^3}

That's it!