# f'(e^2(2x-1))

• Jul 27th 2013, 11:20 AM
Boo
f'(e^2(2x-1))
Dear All!

$\displaystyle f(x)=e^{2x-1}$

The result should be $\displaystyle 2e^{2x-1}$

Can someoe help?
Many thanks!
• Jul 27th 2013, 12:08 PM
Plato
Re: f'(e^2(2x-1))
Quote:

Originally Posted by Boo
$\displaystyle f(x)=e^{2x-1}$
The result should be $\displaystyle 2e^{2x-1}$

If $\displaystyle g$ is a diffentiable function and $\displaystyle y=e^{g(x)}$ then $\displaystyle y'=g'(x)e^{g(x)}$.
• Jul 28th 2013, 06:30 AM
Boo
Re: f'(e^2(2x-1))
Hello
What rule is that?\
Many thanks!!!
• Jul 28th 2013, 07:01 AM
muddywaters
Re: f'(e^2(2x-1))
you have probably previously learned how to differentiate $\displaystyle y=(x^3+1)^3$? it will be $\displaystyle y'= 3x^2 \cdot 3(x^3+1)^2$
try using the same idea to differentiate the question u are doing now (both $\displaystyle y=(x^3+1)^3$ and $\displaystyle f(x)=e^2^x^-^1$are composite functions)
• Jul 28th 2013, 06:48 PM
wondering
Re: f'(e^2(2x-1))
I don't know if this way is easier to understand-

$\displaystyle f(x) = e^{2x-1}$

let u = 2x-1

$\displaystyle f(x) = e^u$

$\displaystyle {f^'}(x) = e^u {u^'} \leftarrow$ that's the chain rule they are talking about

where $\displaystyle {u^'} = 2$ so

$\displaystyle {f^'}(x) = 2e^{2x-1}$
• Jul 31st 2013, 08:00 AM
Boo
Re: f'(e^2(2x-1))
That's it!THANKS!!!!