# Thread: Derivitive with the power rule

1. ## Derivitive with the power rule

Can any one help me solve d/dx$\displaystyle e^(3x^2+8x)$

my answer of$\displaystyle e^(3x^2+8x)(6x+8)(3x^2+8x)+e^(3x^2+8x)(6x+8)$ is not correct.

2. Hello,
Originally Posted by skyslimit
Can any one help me solve d/dx$\displaystyle e^(3x^2+8x)$

my answer of$\displaystyle e^(3x^2+8x)(6x+8)(3x^2+8x)+e^(3x^2+8x)(6x+8)$ is not correct.
Let $\displaystyle u(x)=3x^2+8x$

So you have to find the derivative of $\displaystyle e^{u(x)}$

Chain rule says : $\displaystyle \Bigg([f(g(x))]'=g'(x)f'(g(x))$. Here, $\displaystyle g(x)=u(x)$ and $\displaystyle f(t)=e^t \Bigg)$ -additional part-

Hence we have that the derivative of $\displaystyle e^{u(x)}$ is $\displaystyle u'(x)e^{u(x)}$

Substitute $\displaystyle u(x)$ and $\displaystyle u'(x)$

3. Thanks, thatd worked. Now For more complicated ones $\displaystyle (x-3)^9e^x$

I get $\displaystyle (x-3)^8[9(e^x)+(x-3)(e^x)]$, but that is still not correct. Any suggestions?

4. Originally Posted by skyslimit
Thanks, thatd worked. Now For more complicated ones $\displaystyle (x-3)^9e^x$

I get $\displaystyle (x-3)^8[9(e^x)+(x-3)(e^x)]$, but that is still not correct. Any suggestions?
Hmmm well this is correct o.O
But the completely factorised form is :

$\displaystyle (x-3)^8 e^x [9+(x-3)]=(x-3)^8 e^x [x+6]$

5. Thanks again )