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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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    http://tutorial.math.lamar.edu/Classes/CalcI/ImplicitDIff.aspx2011-11-14
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    You say the definition - but do you really need to do that here? That seems like an unlikely problem. Perhaps just use implicit differentiation?2011-11-14
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    @mixedmath Is it however doable using the def?2011-11-14
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    @Andrew: it is doable in the sense that we have these general proofs for implicit differentiation, the product rule, and the chain rule, and all these rely ultimately only on the definition. So one could, if absolutely necessary, trace these general rules back the definition. However, one would also need to employ some sort of additional argument, likely geometric, for why the derivative of $\cos$ is $-\sin$, as that is not immediate from the definition alone.2011-11-14
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    @mixedmath: Why can't we derive cos' = -sin from the definition (and some other results)?2011-11-14
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    It's possible, and I bet the (some other results) will be geometric.2011-11-14
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    I answered this in your question http://math.stackexchange.com/questions/82061/why-should-i-go-on-and-differentiate-this/82063#82063 three hours ago. Please do not duplicate.2011-11-14

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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 $y'$.