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Thread: derivative of 2(2^x)

  1. #1
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    derivative of 2(2^x)

    Would it be $\displaystyle 0(2^xln2)$ or $\displaystyle 2(2^xln2)$

    It is so confusing....
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by cammywhite View Post
    Would it be $\displaystyle 0(2^xln2)$ or $\displaystyle 2(2^xln2)$

    It is so confusing....
    $\displaystyle 2(2^xln2)$

    $\displaystyle \frac{d(2^{x+1})}{dx} = \frac{d(2(2^x))}{dx}$

    Using Chain Rule:

    $\displaystyle = \frac{2^xd(2)}{~dx} + \frac{ 2d(2^x)}{~dx} $

    = 0 +Ans
    Differntiation of constant is 0 and that of the second term is your answer
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    did you mean to type $\displaystyle 2^{2^x \ln 2}$ ?? if so, you need the chain rule. if you actually meant $\displaystyle 2(2^x \ln 2)$, then no chain rule necessary. just note that $\displaystyle 2 \ln 2$ is a constant, and so we just have to worry about the derivative of $\displaystyle 2^x$, hence

    $\displaystyle \frac d{dx} 2(2^x \ln 2) = 2 (\ln 2^2) \cdot 2^x = 2^{x + 1} (\ln 2)^2$
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