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Thread: Differentiating a exponential function.

  1. #1
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    Differentiating a exponential function.

    Differentiate (w.r.t x ) $\displaystyle e^{2logx + 3x}$

    I tried to solve it as follows:-

    $\displaystyle e^{2logx} + e^{3x}$


    $\displaystyle \frac{\mbox{d}}{\mbox{dx}}(e^{2logx}) + \frac{\mbox{d}}{\mbox{dx}}(e^{3x}) $

    Using chain rule my answer is:-

    $\displaystyle e^{2logx}. \frac{\ 2}{x} + e^{3x}.3 $

    Is it correct.? The actual answer is different.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by vivek_master146 View Post
    Is it correct.? The actual answer is different.

    You can simplify:

    $\displaystyle e^{2\log x}=e^{\log x^2}=x^2$ .

    and your function is:

    $\displaystyle f(x)=x^2e^{3x}$

    not

    $\displaystyle f(x)=x^2+e^{3x}$


    Fernando Revilla
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    Quote Originally Posted by FernandoRevilla View Post
    You can simplify:

    $\displaystyle e^{2\log x}=e^{\log x^2}=x^2$ .

    and your function is:

    $\displaystyle f(x)=x^2e^{3x}$

    not

    $\displaystyle f(x)=x^2+e^{3x}$
    Thanks for reply. The first line totally went over my head. How did you derive $\displaystyle x^2$.

    One more thing:- $\displaystyle e^{a +b} = e^a + e^b \ OR \ e^a.e^b$

    Sorry for simple questions. I am directly studying graduation level maths without studying higher secondary maths.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by vivek_master146 View Post
    Thanks for reply. The first line totally went over my head. How did you derive $\displaystyle x^2$.

    By definition of natural logarithm: $\displaystyle e^{\log A}=A$


    One more thing:- $\displaystyle e^{a +b} = e^a + e^b \ OR \ e^a.e^b$

    The second one.


    Fernando Revilla
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  5. #5
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    You really should learn the "laws of exponents" better before attempting calculus.

    $\displaystyle e^{a+ b}= e^ae^b$, not $\displaystyle e^a+ e^b$,
    $\displaystyle (e^a)^b= e^{ab}$,
    and, since $\displaystyle e^x$ and ln(x) are "inverse" functions,
    $\displaystyle e^{ln(x)}= x$
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