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Well, another problem here I don't get.

From what I know, sin(x)'=cos(x) right?

Well, here is a problem:

Find $\frac{d}{dx}(y\cos(\frac{y}{x^4}))$ ?

Let $u=y\cos(\frac{y}{x^4})$ and $s=y$, $v=\cos(\frac{y}{x^4})$

$$u'=v\frac{ds}{dx}+s \frac{dv}{dx}$$

$$u'=\cos\frac{y}{x^4} \frac{dy}{dx} + y \frac{d}{dx}(\cos\frac{y}{x^4})$$

Now the question, isn't $\frac{d}{dx}(\cos\frac{y}{x^4})=-\sin(\frac{y}{x^4})$ and here it's where it all ends? Why is the teacher going on and differentiates what's in the parentheses?

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Your teacher is correct. S/he is using the chain rule for differentiating the composition of two functions. – Srivatsan Nov 14 '11 at 17:57
So, without it, can't I stop here? – Andrew Nov 14 '11 at 17:57
No, $\sin x$ is not equal to $\cos x$. It's the derivative of $\sin x$ that's equal to $\cos x$. – Michael Hardy Nov 14 '11 at 17:58
@MichaelHardy yea, right. it was just a type – Andrew Nov 14 '11 at 17:59
up vote 1 down vote accepted

No, $\sin x \ne \cos x$

For $\frac{d}{dx}(y*\cos(\frac{y}{x^4}))$ you should be using the product rule and chain rules. It looks like you are integrating by parts. The chain rule is what forces you to differentiate the $\frac{y}{x^4}$



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ok, I got it. It's implicit diff. That's why I have to apply the Chain Rule. – Andrew Nov 14 '11 at 20:28

Take a simpler example, say $\cos(x^2)$.

The derivative is just the slope of the tangent.

Now, going from $\cos(x)$ to $\cos(x^2)$ means that you take the graph of the cosine and stretch it horizontally for large $x$ and squeeze it together for small $x$.

If the tangent has not been horizontal, this stretching and squeezing will change the slope of the tangent, and how far you have stretched or squeezed depends on the rate of change of $x^2$, which explains why you have to multiply by its derivative.

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