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Math Help - Differentiating natural log function

  1. #1
    Member maxpancho's Avatar
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    Differentiating natural log function

    $f(x)=a^x=(a)^x=(e^{\ln a})^x=e^{x \ln a}$

    $f'(x)=$?

    I don't quite understand how the Chain Rule is applied here and what is the inner\outer function? Could someone elaborate?



    PS Oops, I meant natural exponential function.
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    Re: Differentiating natural log function

    Quote Originally Posted by maxpancho View Post
    $f(x)=a^x=(a)^x=(e^{\ln a})^x=e^{x \ln a}$

    $f'(x)=$?

    I don't quite understand how the Chain Rule is applied here and what is the inner\outer function? Could someone elaborate?



    PS Oops, I meant natural exponential function.
    I like dy/dx notation and substitution whenever the chain rule comes up.

    $y = a^x = e^{xln(a)}.$

    $u = xln(a) \implies \dfrac{du}{dx} = ln(a).$

    $And\ u = x * ln(a) \implies y = e^u \implies \dfrac{dy}{du} = e^u = e^{xln(a)} = a^x \implies \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} \implies$

    $\dfrac{dy}{dx} = a^x * ln(a).$
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    Re: Differentiating natural log function

    I don't understand how the derivative of $x \ln a$ and $e^{x \ln a}$ are found..

    $x \ln a$ - product rule?

    $e^{x \ln a}$ ?
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    Re: Differentiating natural log function

    Quote Originally Posted by maxpancho View Post
    I don't understand how the derivative of $x \ln a$ and $e^{x \ln a}$ are found..

    $x \ln a$ - product rule?

    $e^{x \ln a}$ ?
    Remember that a is a constant. So
    \frac{d}{dx} x~ln(a) = \frac{d}{dx}  x~( \text{constant } ) = ( \text{constant } ) = ln(a)

    The second is of the form e^{x~ln(a)} = e^{( \text{constant } )x}. How do you find the derivative of this?

    -Dan
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    Re: Differentiating natural log function

    To differentiate [tex]y= e^{ax}[tex], let u= ax so [tex]\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}[tex]
    (That's the chain rule.)

    \frac{dy}{du}= e^u and \frac{du}{dx}= a so \frac{dy}{dx}= (e^u)(a)= ae^{ax}
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    Re: Differentiating natural log function

    You mention inner and outer functions. I assume you mean the chain rule (with appropriate assumptions) for the derivative of $f(g(x))$ is $f^{\prime}(g(x))g^{\prime}(x)$. Remember an alternate notation for $e^x$ is $\exp(x)$. So $a^x=e^{x\ln(a)}=\exp(x\ln(a))$. Let $f(x)=\exp(x)$ and $g(x)=x\ln(a)$. So the derivative is $f^{\prime}(g(x))g^{\prime}(x)=\exp(x\ln(a))\cdot \ln(a)=a^x\ln(a)$
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  7. #7
    Member maxpancho's Avatar
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    Re: Differentiating natural log function

    So do we leave e intact because e to any imaginable power has the same derivative?
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    Re: Differentiating natural log function

    Or, rather e depends on it's convoluted exponent same as $e^x$ depends on x, when we differentiate... so we just rewrite e with its convoluted exponent, much the same as we would do if we defferentiated $e^x$ over x?


    $w=x^2+3$

    $\dfrac{d}{dw} e^{x^2+3}=e^{x^2+3}$

    because

    $\dfrac{d}{dx} e^x=e^x$
    Last edited by maxpancho; August 8th 2014 at 10:52 AM.
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    Re: Differentiating natural log function

    I have no idea what a convoluted exponent is. Perhaps it is what happens to an exponent that has been out drinking with contortionists.

    But yes. (Note: I truly wish that, in beginning explanations. math texts would not use the same letter to refer to different variables.)

    $y = e^x \implies \dfrac{dy}{dx} = e^x.$ That is a basic derivative, a general rule, just like the addition, product, and power rules.

    $\therefore w = u^2 + 3\ and\ z = e^{(u^2 + 3)} \implies z = e^w\ and \dfrac{dz}{dw} = e^w = e^{(u^2 + 3)}.$ We applied the general rule about powers of e.

    $But\ w = u^2 + 3 \implies \dfrac{dw}{du} = 2u.$ We applied the general rules about the derivatives of sums and the derivatives of a variable to a power.

    $\therefore \dfrac{dz}{du} = \dfrac{dz}{dw} * \dfrac{dw}{du} = e^{(u^2 + 3)} * 2u = 2ue^{(u^2 + 3)}.$ We applied the chain rule.
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  10. #10
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    Re: Differentiating natural log function

    The chain rule:

    $(f \circ g)'(u) = f'(g(u))\cdot g'(u)$

    For the function:

    $h(x) = a^x = (e^{\log(a)})^x = e^{(\log(a))x}$,

    let $f(x) = e^x$ and let $g(x) = (\log(a))x$ <--$g(x)$ is just a constant (in this case, the constant is $\log(a)$) times $x$, that is $g(x) = cx$, where $c = \log(a)$.

    Then $h = f \circ g$.

    Applying the chain rule, we have $f'(x) = e^{x}$, and $g'(x) = c = \log(a)$.

    So at the point $x = u$, we have $g(u) = (\log(a))u$, and $f'(g(u)) = e^{g(u)} = e^{(\log(a))u}$, and we have $g'(u) = c = \log(a)$.

    To get the derivative of $h = f \circ g$ at the point $u$, we multiply these two numbers together:

    $h'(u) = (e^{(\log(a))u})\cdot \log(a)$.

    We can simplify this a bit, since (reversing the steps we took earlier), $e^{(\log(a))u} = (e^{\log(a)})^u = a^u$, so that $h'(u) = a^u \cdot \log(a)$.

    So the derivative of $h$, as a function of $x$, is given by:

    $h'(x) = a^x\cdot\log(a)$.

    Note that gives us the same answer as before, when $a = e$:

    Writing $e^x$ as $\exp(x)$, we have:

    $\exp'(x) = e^x\cdot \log(e) = e^x\cdot 1 = e^x = \exp(x)$.
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