# Thread: More Chain Rule Help

1. ## More Chain Rule Help

Every time I feel like I understand it, something new comes up and proves me wrong. I have been given an example in my text and I did the question before looking at the answer. I don't understand why they did it the way they did and I feel like my answer is correct. Clearly I do not understand how to do this right. I was hoping someone could clear things up for me.
$\displaystyle f(x)=3^{x^2+1}-x^2\\f\prime(x)=3^{x^2+1}\cdot ln(3)\cdot 2x-(2x)$
$\displaystyle f(x)=3^{x^2+1}-x^2\\f\prime(x)=(x^2+1)\prime3^{x^2+1}\cdot ln(3)-2x\\=(2x)3^{x^2+1}\cdot ln(3)-2x$

Also, the chain rule I have been given for exponential functions is this:
$\displaystyle f(x)=a^{g(x)}$ where a is a positive constant, is given by $\displaystyle f\prime(x)=a^{g(x)} ln(a) g\prime(x)$, which I paired with the difference rule.

So I guess what I'm really trying to figure out here, is why they put $\displaystyle g\prime(x)$ in front?

2. ## Re: More Chain Rule Help

Originally Posted by whit221
Every time I feel like I understand it, something new comes up and proves me wrong. I have been given an example in my text and I did the question before looking at the answer. I don't understand why they did it the way they did and I feel like my answer is correct. Clearly I do not understand how to do this right. I was hoping someone could clear things up for me.
$\displaystyle f(x)=3^{x^2+1}-x^2\\f\prime(x)=3^{x^2+1}\cdot ln(3)\cdot 2x-(2x)$
$\displaystyle f(x)=3^{x^2+1}-x^2\\f\prime(x)=(x^2+1)\prime3^{x^2+1}\cdot ln(3)-2x\\=(2x)3^{x^2+1}\cdot ln(3)-2x$

Also, the chain rule I have been given for exponential functions is this:
$\displaystyle f(x)=a^{g(x)}$ where a is a positive constant, is given by $\displaystyle f\prime(x)=a^{g(x)} ln(a) g\prime(x)$, which I paired with the difference rule.

So I guess what I'm really trying to figure out here, is why they put $\displaystyle g\prime(x)$ in front?
Your answer: $f\prime(x)= 3^{x^2+1}\cdot ln(3)\cdot 2x-(2x).$

Book answer: $f\prime(x)= (2x)3^{x^2+1}\cdot ln(3)-2x.$

Those two answers mean exactly the same thing: multiplication is commutative.

3. ## Re: More Chain Rule Help

I guess I was stressing over nothing. Thanks! ... just not sure why they needed to rearrange it.