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Thread: f'(e^2(2x-1))

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

    Quote Originally Posted by Boo View Post
    $\displaystyle f(x)=e^{2x-1}$
    The result should be $\displaystyle 2e^{2x-1}$
    If $\displaystyle g$ is a diffentiable function and $\displaystyle y=e^{g(x)}$ then $\displaystyle y'=g'(x)e^{g(x)}$.
    Thanks from topsquark
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  3. #3
    Boo
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    Re: f'(e^2(2x-1))

    Hello
    Please,I don't understand this.
    What rule is that?\
    Many thanks!!!
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  4. #4
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    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)
    Thanks from topsquark
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  5. #5
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    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

    $\displaystyle {f^'}(x) = 2e^{2x-1}$
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  6. #6
    Boo
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    Re: f'(e^2(2x-1))

    That's it!THANKS!!!!
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