Question

derivative of f(x)=csc(pi*x^2). Can't you just start w/ f'(x)= -csc(pi*x^2)*cot(pi*x^2) ?

Answer 1

\[\sf \csc(\pi x^2)~??\]

Answer 2

yep

Answer 3

Are you familiar with chain rule? derivative of the outside times the dervative of the inside. In otherwords, you have: \(\sf \color{red}{\frac{d}{dx}outside(inside) \times \frac{d}{dx}(inside)}\)

Answer 4

Remembr that \(\pi\) is a constant.

Answer 5

I mean, isn't the following true? The derivative of:
\[\csc(x) = -\csc(x)\cot(x)\]

Answer 6

I'm familiar w/ the chain rule but it's getting a little fuzzy w/ the trig functions

Answer 7

Yes, thats correct. So you would have\(\sf \color{blue}{-cot(\pi x^2)(csc(\pi x^2)}\)\(\sf \color{red}{\times \frac{d}{dx}(\pi x^2)}\) = ____________________

Answer 8

ok, so I can't just take the derivative like I did 3 lines up then? It's still a chain rule type relationship w/ these?

Answer 9

I don't know what you mean. Unless you're asked to find \(\sf \color{purple}{f''(x)}\), what I showed up there, is all you need to do. And of course, don't forget to take the deriviative of the inside function, as I did not finish the problem. I left it for you to finish.

Answer 10

ok, so take a more generic approach.
let \[f(x)=\csc(x)\]
\[f \prime(x)=-\csc(x)\cot(x)\]
or is:
\[f \prime(x)=-\csc(x)\cot(x)*(x)'\]

Sorry, I'm just learning the syntax.

Answer 11

Yes.

Answer 12

Which one? The first or the second solution?

Answer 13

The second one.

Answer 14

ok, somehow I missed that in my notes. I thought it was just the first.

Answer 15

According to the chain rule, the second one is correct, but you don't need to show that since you know that \(\sf \frac{d}{dx}~x = 1\)

Answer 16

oh, I see now... in class they were just using sin(x) (etc.) in the examples. They didn't do things like sin(pi*x) and elaborate on that.

Answer 17

That makes sense now

Answer 18

Thanks so much. You rock and this site is pretty baller!