# How to differentiate $\cos(y/x^4)$? [duplicate]

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.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new 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 math.stackexchange.com/questions/82061/… three hours ago. Please do not duplicate. –  Ross Millikan Nov 14 '11 at 21:19

## 1 Answer

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 $y'$.

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