# Differentiation

• Apr 18th 2010, 02:59 PM
mike21
Differentiation
need help differentiating (6e^2x-x)^3

also finding the first and second derivatives for f(x)=1-2x over e^x

I used the general power rule to get -2(e^x)-e^x(1-2x) over (ex)^2, but I dont know if thats right and don't know where to go from there to get the second derivative.

Also for this problem e^(x^2+9)times e^(6x)=1 can I just drop the base's because there the same? how would you solve this?

Any help is greatly appreciated thanks.
• Apr 18th 2010, 03:50 PM
skeeter
Quote:

Originally Posted by mike21
need help differentiating (6e^2x-x)^3

chain rule ...

$\displaystyle \textcolor{red}{3(6e^{2x} - x)^2 \cdot (12e^{2x} - 1)}$

also finding the first and second derivatives for f(x)=1-2x over e^x

quotient rule ...

$\displaystyle \textcolor{red}{f'(x) = \frac{e^x(-2) - (1-2x)e^x}{e^{2x}}}$

$\displaystyle \textcolor{red}{f'(x) = \frac{e^x[-2 - (1-2x)]}{e^{2x}}}$

$\displaystyle \textcolor{red}{f'(x) = \frac{e^x(2x-3)}{e^{2x}} = \frac{2x-3}{e^x}}$

now do the quotient rule again

Also for this problem e^(x^2+9)times e^(6x)=1 can I just drop the base's because there the same? how would you solve this?

$\displaystyle \textcolor{red}{e^{x^2+9} \cdot e^{6x} = e^{x^2+6x+9} = e^{(x+3)^2} = 1}$

$\displaystyle \textcolor{red}{(x+3)^2 = 0}$

...