Question

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!

Answer 1

`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!