# Thread: Help with Chain Rule!!!

1. ## Help with Chain Rule!!!

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

2. 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)$

3. 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 $e^{g(x)}\cdot g'(x)$.

4. 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.

5. 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.

6. 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!

7. No problem. You weren't nasty at all.
I've even been called names here, right mr fancy.