Dear All!

Please, I can't solve:

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

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

Can someoe help?

Many thanks!

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

Please, I can't solve:

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

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

Can someoe help?

Many thanks! - Jul 27th 2013, 12:08 PMPlatoRe: f'(e^2(2x-1))
- Jul 28th 2013, 06:30 AMBooRe: f'(e^2(2x-1))
Hello

Please,I don't understand this.

What rule is that?\

Many thanks!!! - Jul 28th 2013, 07:01 AMmuddywatersRe: 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 PMwonderingRe: 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 AMBooRe: f'(e^2(2x-1))
That's it!THANKS!!!!