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Thread: Differentiation of 2^n with respect to n

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    Differentiation of 2^n with respect to n

    i was searching on the net on how to solve the question listed in the subject. The answer was given but with no proper explanation. Can someone roughly guide me on how to get the answer.

    d/dn(2^n) = ln2 x 2^n ( how to get this answer )
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    Quote Originally Posted by tester85 View Post
    i was searching on the net on how to solve the question listed in the subject. The answer was given but with no proper explanation. Can someone roughly guide me on how to get the answer.

    d/dn(2^n) = ln2 x 2^n ( how to get this answer )
    If you are convinced that the derivative of \ln{(x)} W.R.T. x is \frac{1}{x}, then consider this:

    y = 2^x \implies \ln{(y)} = x\ln{2}

    Implicitly differentiating, we get:
    \frac{1}{y}y' = \ln{2} \implies y' = y\ln{2} = 2^x \ln{2}
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    Quote Originally Posted by tester85 View Post
    i was searching on the net on how to solve the question listed in the subject. The answer was given but with no proper explanation. Can someone roughly guide me on how to get the answer.

    d/dn(2^n) = ln2 x 2^n ( how to get this answer )
    Let y = 2^n

    Then \ln{y} = \ln{2^n}

    \ln{y} = n\ln{2}

    y = e^{n\ln{2}}.


    Now, to find this derivative, use the chain rule. Let u = n\ln{2}, so y = e^u.

    \frac{du}{dx} = \ln{2}

    \frac{dy}{du} = e^u = e^{n\ln{2}} = e^{\ln{2^n}} = 2^n.


    \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}

     = \ln{2}\,2^n.
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