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Thread: Derivative of an exponential when the exponent is a function

  1. #1
    Member
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    Derivative of an exponential when the exponent is a function

    what is the value of
    $\displaystyle
    \frac{dy}{dx}
    $
    if
    $\displaystyle
    y=2^{\sin (2^{\sin(2^x)})}
    $
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  2. #2
    Senior Member Twig's Avatar
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    Use the fact that $\displaystyle a^{f(x)}=e^{ln(a)^{f(x)}}=e^{f(x)\cdot ln(a)} $
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  3. #3
    Super Member Random Variable's Avatar
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    $\displaystyle y = 2^{\sin 2^{sin2^{x}}} $

    let $\displaystyle u= 2^{x} $

    then $\displaystyle y = 2^{\sin 2^{\sin u}} $

    let $\displaystyle v = \sin u $

    then $\displaystyle y = 2^{\sin 2^{v}} $

    let $\displaystyle w = 2^v $

    then $\displaystyle y = 2^{\sin w} $

    finally let $\displaystyle z = \sin w $

    then $\displaystyle y = 2^{z} $

    $\displaystyle \frac {dy}{dx} = \frac {dy}{dz} \frac {dz}{dw} \frac {dw}{dv} \frac {dv}{du} \frac {du}{dx} $

    $\displaystyle \frac {dy}{dx} = \ln(2) 2^{z} \cos(w) ln(2)2^{v} \cos(u) ln(2) 2^x $

    $\displaystyle = \ln(2) 2^{\sin w} \cos(2^{v}) ln(2)2^{\sin u} \cos(2^{x}) \ln(2) 2^{x} $

    $\displaystyle = \ln(2) 2^{\sin 2^{v}} \cos(2^{\sin u}) \ln(2)2^{\sin 2^{x}} \cos(2^{x}) \ln(2) 2^{x} $

    $\displaystyle = \ln(2) 2^{\sin 2^{\sin 2^x}} \cos(2^{\sin 2^{x}}) \ln(2)2^{\sin 2^{x}} \cos(2^{x}) \ln(2) 2^{x} $
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