Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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
1  
@MichaelHardy yea, right. it was just a type –  Andrew Nov 14 '11 at 17:59

2 Answers 2

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 $\frac{y}{x^4}$

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

$\frac{d}{dx}cos(\frac{y}{x^4})=-\sin(\frac{y}{x^4})\frac{d}{dx}(\frac{y}{x^4})=-\sin(\frac{y}{x^4})\left(\frac{dy}{dx}\frac{1}{x^4}-4yx^{-5}\right)$

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

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.