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

Possible Duplicate:
Why should I go on and differentiate this?

Some help please, I know how to differentiate $\cos x$ but what about $$\frac{d}{dx}\cos\left(\frac{y}{x^4}\right)?$$ I tried to plug it into the definition but with no success.

share|cite|improve this question

marked as duplicate by mixedmath, Srivatsan, t.b., Gerry Myerson, Henning Makholm Nov 15 '11 at 1:23

This question was marked as an exact duplicate of an existing question.

You say the definition - but do you really need to do that here? That seems like an unlikely problem. Perhaps just use implicit differentiation? – mixedmath Nov 14 '11 at 20:28
@mixedmath Is it however doable using the def? – Andrew Nov 14 '11 at 20:30
@mixedmath: Why can't we derive cos' = -sin from the definition (and some other results)? – The Chaz 2.0 Nov 14 '11 at 20:43
I answered this in your question… three hours ago. Please do not duplicate. – Ross Millikan Nov 14 '11 at 21:19
up vote 0 down vote accepted

NOTE: I see now that this is the more or less same answer as given by Ross here, and that the OP asked the question twice.

I will explain how to do it without using the definition of a derivative.

We are to differentiate $\cos \left( \dfrac{y}{x^4} \right)$. Then the derivative of $\cos$ is $-\sin$, so we get $-\sin \left( \dfrac{y}{x^4} \right) \cdot \left( \dfrac{y}{x^4} \right)'$. What is $\left( \dfrac{y}{x^4} \right)'$?

$\left( \dfrac{y}{x^4} \right)' = y \cdot \dfrac{-4}{x^5} + y' \cdot \dfrac{1}{x^4}$

Of course, knowing nothing more about $y$, we cannot simplify $y'$.

share|cite|improve this answer

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