Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello everyone how would I solve the following derivative.


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

So would I do


As my derivative.

share|cite|improve this question
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
up vote 7 down vote accepted

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): $$ \frac{d}{dx} \cot(2x) = -\csc^{\color{red} 2}(2x)\frac{d}{dx}(2x). $$

share|cite|improve this answer
+1 for not providing the entire solution. – JavaMan Feb 16 '13 at 16:26
On the second term do you mean the to use the chain rule? – Fernando Martinez Feb 16 '13 at 16:29
@FernandoMartinez: That is exactly right. – Thomas Feb 16 '13 at 16:30
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.