# Trigonometric derivative?

Hello everyone how would I solve the following derivative.

$f(x)=5x^3\tan(x)+\cot(2x)$

I know the derivative of $\tan(x)$ is $\sec^2(x)$

So would I do

$15x^2\sec^2(x)-\csc(2x)$

As my derivative.

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Product rule... –  N. S. Feb 16 '13 at 16:24
yes in the first term I think.I have to use it. –  Fernando Martinez Feb 16 '13 at 16:25
Also, as a note on the vocabulary, you do not "solve" a derivative (there is no equation), but rather "find" the derivative. –  JavaMan Feb 16 '13 at 16:26

Notice that you have a product of two functions in the first term: $$5x^3\tan(x).$$ So you need to use the product rule. You get (for the first term only) $$\frac{d}{dx} 5x^3\tan(x) = \left[\frac{d}{dx}5x^3\right]\tan(x) + 5x^3\frac{d}{dx}\tan(x).$$
Also for the second term $$\cot(2x)$$ you need to multiply by the derivative of the "inner function" $2x$ (using the Chain Rule here): $$\frac{d}{dx} \cot(2x) = -\csc^{\color{red} 2}(2x)\frac{d}{dx}(2x).$$
So for my final answer I got $15x^2\tan(x)+5x^3\sec^2(x)-\csc^2(2x)(2)$ –  Fernando Martinez Feb 16 '13 at 16:32
As a matter of style, I would prefer $2\csc^2(2x)$ towards the end of your expression. That's because, particularly if you do further processing in a problem, it might be easy to misread $\csc^2(2x)(2)$ as meaning $\csc^2(4x)$. –  André Nicolas Feb 16 '13 at 17:56