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

Dear All!

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

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

Can someoe help?
Many thanks!

2. ## Re: f'(e^2(2x-1))

Originally Posted by Boo
$f(x)=e^{2x-1}$
The result should be $2e^{2x-1}$
If $g$ is a diffentiable function and $y=e^{g(x)}$ then $y'=g'(x)e^{g(x)}$.

3. ## Re: f'(e^2(2x-1))

Hello
What rule is that?\
Many thanks!!!

4. ## Re: f'(e^2(2x-1))

you have probably previously learned how to differentiate $y=(x^3+1)^3$? it will be $y'= 3x^2 \cdot 3(x^3+1)^2$
try using the same idea to differentiate the question u are doing now (both $y=(x^3+1)^3$ and $f(x)=e^2^x^-^1$are composite functions)

5. ## Re: f'(e^2(2x-1))

I don't know if this way is easier to understand-

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

let u = 2x-1

$f(x) = e^u$

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

where ${u^'} = 2$ so

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

6. ## Re: f'(e^2(2x-1))

That's it!THANKS!!!!