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Math Help - Chain Rule Differentiation

  1. #1
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    Chain Rule Differentiation

    2^(Πsinx)


    Seems like I'm having difficulty understanding the a^x = a*ln(a) rule...
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  2. #2
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    Quote Originally Posted by Neversh View Post
    2^(Πsinx)


    Seems like I'm having difficulty understanding the \color{blue}a^x = a^x*ln(a) rule...
    Hi Neversh,


    \frac{d}{dx}2^{{\pi}sinx}=\frac{du}{du}\frac{d}{dx  }2^{{\pi}sinx}

    u={\pi}sinx

    a=2

    \frac{du}{dx}\frac{d}{du}2^u=({\pi}cosx)2^uln2

    =({\pi}cosx)2^{{\pi}sinx}ln2
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