Help with Chain Rule!!!

**sk8erboyla2004** · March 12th 2009, 09:38 PM

I have the very basic of the chain rule understood however the functions

with powers that contain equation confuse the hell out of me I hope to get some prompt and excellent health let me show an example

$f(x)=e^{(1+3t)^{2}}$

and the chain rule is

$ f\prime(g(x)) * g\prime(x) $

and the answer i was shown and the stops shown dont make sense from what i think i understand

$f\prime(x)=(e^{1+3t^{2}})(2)(1+3x)3$

I can understand $f\prime(x)=(e^{1+3t^{2}})$ to
$f\prime(g(x))$ and $(2)(1+3x)$ to be $g\prime$ however why would you derive $(1+3x)$ as you did all steps of the chain rule why would you need a further derivation

pretty much anything to power containing a constant confuses me

hope this posts get a reply unlike my other ones

**mollymcf2009** · March 12th 2009, 10:03 PM

Originally Posted by sk8erboyla2004

I have the very basic of the chain rule understood however the functions

with powers that contain equation confuse the hell out of me I hope to get some prompt and excellent health let me show an example

$f(x)=e^{(1+3t)^{2}}$

and the chain rule is

$ f\prime(g(x)) * g\prime(x) $

and the answer i was shown and the stops shown dont make sense from what i think i understand

$f\prime(x)=(e^{1+3t^{2}})(2)(1+3x)3$

I can understand $f\prime(x)=(e^{1+3t^{2}})$ to
$f\prime(g(x))$ and $(2)(1+3x)$ to be $g\prime$ however why would you derive $(1+3x)$ as you did all steps of the chain rule why would you need a further derivation

pretty much anything to power containing a constant confuses me

hope this posts get a reply unlike my other ones

Chain rule twice:

$f(x)=e^{(1+3t)^{2}}$

Let $u=(1+3t)^2$
$du =2(1+3t) \cdot 3$

$f'(x) = e^u du$

$= e^{{1+3t}^2} \cdot 2(1+3t) \cdot 3$
$ = 6e^{{1+3t}^2} \cdot (1+3t)$

**matheagle** · March 12th 2009, 10:11 PM

Originally Posted by mollymcf2009

Chain rule twice:

$f(x)=e^{(1+3t)^{2}}$

Let $u=(1+3t)^2$
$du =2(1+3t) \cdot 3$

$f'(x) = e^u du$

$= e^{(1+3t)^2} \cdot 2(1+3t) \cdot 3$ Mr F edit: Excuse me for butting in. matheagle has corrected this line (and the next).The original line was $= e^{{1+3t}^2} \cdot 2(1+3t) \cdot 3$ . (So what matheagle has said is perfectly correct).

$ = 6e^{(1+3t)^2} \cdot (1+3t)$

This is what you meant $6(1+3t)e^{(1+3t)^2}$ .

The derivative of $e^{g(x)}$ is $e^{g(x)}\cdot g'(x)$ .

**mollymcf2009** · March 12th 2009, 10:15 PM

Originally Posted by matheagle

This is what you meant $6(1+3t)e^{(1+3t)^2}$ .

The derivative of $e^{g(x)}$ is $e^{g(x)}\cdot g'(x)$ .

That's what I did have. Same thing.

**matheagle** · March 12th 2009, 10:18 PM

Originally Posted by mollymcf2009

That's what I did have. Same thing.

You meant that, but look at what you typed.
You have a typo. It's a TeX error. You used {} instead of (), making the exponent of e, accidentally $1+3t^2$ instead of $(1+3t)^2$ .
I know you know what you're doing, but the typo will cause confusion. That's why I pointed it out.

**mollymcf2009** · March 13th 2009, 06:56 PM

Originally Posted by matheagle

You meant that, but look at what you typed.
You have a typo. It's a TeX error. You used {} instead of (), making the exponent of e, accidentally $1+3t^2$ instead of $(1+3t)^2$ .
I know you know what you're doing, but the typo will cause confusion. That's why I pointed it out.

You're totally right. When you quoted me, I didn't realize that you had corrected within my quote. I just read it like it was what I had put, so I couldn't understand what you were talking about. Thanks for catching that. Those parentheses MUST be there. Thanks and sorry for being short with you, I was SO tired!! Have a great weekend!

**matheagle** · March 13th 2009, 08:13 PM

No problem. You weren't nasty at all.
I've even been called names here, right mr fancy.
I debated about fixing your typo or not within the quote.
In the end I decided to type it correctly.
My concern here is that students will get confused over typos like that.

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