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Thread: Derivitive with the power rule

  1. #1
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    Derivitive with the power rule

    Can any one help me solve d/dx$\displaystyle e^(3x^2+8x) $

    my answer of$\displaystyle e^(3x^2+8x)(6x+8)(3x^2+8x)+e^(3x^2+8x)(6x+8) $ is not correct.
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  2. #2
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    Hello,
    Quote Originally Posted by skyslimit View Post
    Can any one help me solve d/dx$\displaystyle e^(3x^2+8x) $

    my answer of$\displaystyle e^(3x^2+8x)(6x+8)(3x^2+8x)+e^(3x^2+8x)(6x+8) $ is not correct.
    Let $\displaystyle u(x)=3x^2+8x$

    So you have to find the derivative of $\displaystyle e^{u(x)}$

    Chain rule says : $\displaystyle \Bigg([f(g(x))]'=g'(x)f'(g(x))$. Here, $\displaystyle g(x)=u(x)$ and $\displaystyle f(t)=e^t \Bigg)$ -additional part-

    Hence we have that the derivative of $\displaystyle e^{u(x)}$ is $\displaystyle u'(x)e^{u(x)}$


    Substitute $\displaystyle u(x)$ and $\displaystyle u'(x)$
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    Thanks, thatd worked. Now For more complicated ones $\displaystyle (x-3)^9e^x$

    I get $\displaystyle (x-3)^8[9(e^x)+(x-3)(e^x)] $, but that is still not correct. Any suggestions?
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  4. #4
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    Quote Originally Posted by skyslimit View Post
    Thanks, thatd worked. Now For more complicated ones $\displaystyle (x-3)^9e^x$

    I get $\displaystyle (x-3)^8[9(e^x)+(x-3)(e^x)] $, but that is still not correct. Any suggestions?
    Hmmm well this is correct o.O
    But the completely factorised form is :

    $\displaystyle (x-3)^8 e^x [9+(x-3)]=(x-3)^8 e^x [x+6]$
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    Thanks again )
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