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

Need some help with the steps in converting the derivatives of the following functions.

  1. derivative of $\cos(\tan(x))$ to $\frac{-\sin(\tan (x))}{\cos^2(x)}$

    I can get $-\sec^2(x) \cdot (\sin(\tan(x))$ using chain rule, but then I am stuck. I guess I just need help on understanding how $\sec^2(x) = \frac{1}{\cos^2(x)}$

  2. derivative of $\sin(x)\tan(x)$ to $\sin(x) + \tan(x) \cdot \sec(x)$

I can get $\cos(x) \cdot \sec^2(x)$ but then I am unsure what to do. Thanks for any help!

share|cite|improve this question
up vote 2 down vote accepted

For your first question, remember that $\sec$ is the reciprocal of $\cos$ by definition.

For the second, you should use the product rule to find the derivative, don't simply take the derivative of each factor. Then use the fact that $\tan{x}=\frac{\sin{x}}{\cos{x}}$ to simplify your expression and get the desired form. Try rewriting every function in terms of $\sin$ and $\cos$ to most easily simplify.

share|cite|improve this answer
Thanks, I see using product rule I get sinx(1/cox^2(x)) + (sinx/cosx)(cosx) which simplifies to sinx + tanxsecx – Finzz Jan 30 '11 at 22:31
@Finzz, exactly, nice job. – yunone Jan 30 '11 at 23:04
  1. $\sec(x)$ is defined to be $\frac{1}{\cos(x)}$, that is the ratio of the length of the hypotenuse and the length of the adjacent side.

  2. Note that $\tan(x)\cdot \sec(x) = \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)} = \sin(x) \cdot \frac{1}{\cos^2(x)}$.

share|cite|improve this answer

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.