# Thread: Derivative and Chain Rule HELP!!!!

1. ## Derivative and Chain Rule HELP!!!!

I've been stressing over these problems all week and my math professor has been always "busy" after school (I know... unreliable). So could anyone please help me with a few of these calculus problems and please explain how you got to the answer? I'm not going to type all of my homework because I also want to work on the rest of the problems so I am sure I am getting it.

Here are the directions and a couple problems:

Use the derivative formulas and the chain rule: if $(fog)(x) = f(g(x)) => (fog)'(x) = f'(g(x)) * g'(x)$ To find the derivative $\frac {dy}{dx}$ of each of the functions below.

10) $y = cos^2(2x)$

11) $y = x* sin(x^2)$

12) $y = \frac{x+sinx}{cosx}$

Could someone out there please help! We have a quiz on this on Friday and I need all the help I could get.

2. Originally Posted by RedSpades
I've been stressing over these problems all week and my math professor has been always "busy" after school (I know... unreliable). So could anyone please help me with a few of these calculus problems and please explain how you got to the answer? I'm not going to type all of my homework because I also want to work on the rest of the problems so I am sure I am getting it.

Here are the directions and a couple problems:

Use the derivative formulas and the chain rule: if $(fog)(x) = f(g(x)) => (fog)'(x) = f'(g(x)) * g'(x)$ To find the derivative $\frac {dy}{dx}$ of each of the functions below.

10) $y = cos^2(2x)$

11) $y = x* sin(x^2)$

12) $y = \frac{x+sinx}{cosx}$

Could someone out there please help! We have a quiz on this on Friday and I need all the help I could get.
10) $y' = 2\cos(2x)*2$

11) $y' = \sin(x^2) + x \cos(x^2) 2x$

12) I'll leave this up to you...

3. Originally Posted by colby2152
10) $y' = 2\cos(2x)*2$

11) $y' = \sin(x^2) + x \cos(x^2) 2x$

12) I'll leave this up to you...
Thanks colby, but I would really appreciate it if you could explain how you got those answers. I would like to study how to answer these problems so I don't depend on this forum to provide me with just the answers. I want to learn this stuff so I won't bother you guys. :P

4. Originally Posted by RedSpades
Thanks colby, but I would really appreciate it if you could explain how you got those answers. I would like to study how to answer these problems so I don't depend on this forum to provide me with just the answers. I want to learn this stuff so I won't bother you guys. :P
Well, you start off deriving one function such as cosine...

$f(x)=\cos(2x)$

$f'(x)=-\sin(2x) * ??$

It's derivative is negative sine. Now, you must follow the chain into the old cosine function and take the derivative of its parameter with respect to x. The derivative of 2x is just 2, so:

$f'(x)=-\sin(2x) * 2$

$f'(x)=-2\sin(2x)$

With a normal parameter trig function such as $f(x)=\sin(x)$, you always practice the chain rule when deriving, but you are not aware of it..

$f'(x)=\cos(x)*1$

One is the derivative of x, but the one simplifies itself because of the identity property!

$f'(x)=\cos(x)$

5. So how would you do problem 12 since it is a divison problem. Wouldn't be such as f(x)/g(x). I'm just confused on those types of problems. Please help me out.