# New Calculus Problem

• Oct 29th 2006, 11:13 AM
Sucker Punch
Sorry to ressurect this, but I have another assignment on similar subject matter. I figured it would be better to ask again her, than make a new thread.

Both of these types I remember covering in class, but can't seem to find them in my notes.

$\displaystyle e^{2x^3}$
That is e to the power of 2 times x to the power of 3.

Do I double up the chain rule here or something?

Also
$\displaystyle sin(tan(x))$

would that just become
$\displaystyle -cos(sec^2(x))$

Or is there more to it?
• Oct 29th 2006, 11:25 AM
CaptainBlack
Don't tag a new question onto the end of an existing thread, it makes it
difficult to identify that a new question has been asked.

RonL
• Oct 29th 2006, 11:29 AM
CaptainBlack
Quote:

Originally Posted by Sucker Punch
Sorry to ressurect this, but I have another assignment on similar subject matter. I figured it would be better to ask again her, than make a new thread.

Both of these types I remember covering in class, but can't seem to find them in my notes.

$\displaystyle e^{2x^3}$
That is e to the power of 2 times x to the power of 3.

Do I double up the chain rule here or something?

Chain rule:

$\displaystyle \frac{d}{dx}f(g(x))=g'(x) f'(g(x))$

In this case $\displaystyle g(x)=2x^3$ and $\displaystyle f(x)=e^x$, so

$\displaystyle g'(x)=6x^2$

and

$\displaystyle f'(x)=e^x$

So:

$\displaystyle \frac{d}{dx}e^{2x^3}=6x^2\ e^{2x^2}$

RonL
• Oct 29th 2006, 11:37 AM
CaptainBlack
Quote:

Originally Posted by Sucker Punch
Also
$\displaystyle sin(tan(x))$

would that just become
$\displaystyle -cos(sec^2(x))$

Or is there more to it?

Chain rule:

$\displaystyle \frac{d}{dx}[\sin(\tan(x))]=\left\{ \frac{d}{dx}[\tan(x)]\right\}\ \cos(\tan(x)) = sec^2(x)\ \cos(\tan(x))$

RonL