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Math Help - Logarithmic differentiation

  1. #1
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    Logarithmic differentiation

    Logarithmic differentiation
    Having some problems with these question....


    f(x)
    and g(x) are two differentiable functions and f(x)>0 for all x.
    Consider h(x) = f(x)^g(x)

    a) Show that ln h(x) = g(x) ln f(x)

    b) Show that the derivate of ln h(x) with respect to x is:
    g(x) ln f(x) + g(x) f(x)/f(x)

    c) Show that the derivate of ln h(x) with respect to x is:
    h(x)/h(x)

    d) Show that h(x) = f(x)^g(x) [g(x) ln f(x) + g(x) f(x)/f(x)]

    e) Use the above formula to find the derivate of h(x)=x^x.
    Identify the functions f(x) and g(x) and their derivatives.


    Thanks for help!! This is killing me...
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  2. #2
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    Quote Originally Posted by Lippi86 View Post
    Logarithmic differentiation
    Having some problems with these question....


    f(x)
    and g(x) are two differentiable functions and f(x)>0 for all x.
    Consider h(x) = f(x)^g(x)

    a) Show that ln h(x) = g(x) ln f(x)

    b) Show that the derivate of ln h(x) with respect to x is:
    g(x) ln f(x) + g(x) f(x)/f(x)

    c) Show that the derivate of ln h(x) with respect to x is:
    h(x)/h(x)

    d) Show that h(x) = f(x)^g(x) [g(x) ln f(x) + g(x) f(x)/f(x)]

    e) Use the above formula to find the derivate of h(x)=x^x.
    Identify the functions f(x) and g(x) and their derivatives.


    Thanks for help!! This is killing me...
    a) h(x) = f(x)^{g(x)}

    \ln{\left[h(x)\right]} = \ln{\left[f(x)^{g(x)}\right]}

    \ln{\left[h(x)\right]} = g(x)\ln{\left[f(x)\right]}.


    b) \frac{d}{dx} \left\{\ln{\left[h(x)\right]}\right\}= \frac{d}{dx}\left\{g(x)\ln{\left[f(x)\right]}\right\}

     = g(x)\,\frac{d}{dx}\left\{\ln{\left[f(x)\right]}\right\} + \ln{\left[f(x)\right]}\,\frac{d}{dx}[g(x)]

     = \frac{g(x)f'(x)}{f(x)} + g'(x)\ln{\left[f(x)\right]}


    c) y = \ln{\left[h(x)\right]}.

    Let u = h(x) so that y = \ln{u}.

    \frac{du}{dx} = h'(x)

    \frac{dy}{du} = \frac{1}{u} = \frac{1}{h(x)}.

    \frac{dy}{dx} = \frac{h'(x)}{h(x)}.


    Therefore \frac{d}{dx}\left\{\ln{\left[h(x)\right]}\right\} = \frac{h'(x)}{h(x)}.


    d) Since \frac{d}{dx}\left\{\ln{\left[h(x)\right]}\right\} = \frac{h'(x)}{h(x)} and \frac{d}{dx} \left\{\ln{\left[h(x)\right]}\right\}= \frac{g(x)f'(x)}{f(x)} + g'(x)\ln{\left[f(x)\right]}

    This means

    \frac{h'(x)}{h(x)} = \frac{g(x)f'(x)}{f(x)} + g'(x)\ln{\left[f(x)\right]}.


    So h'(x) = h(x)\left\{\frac{g(x)f'(x)}{f(x)} + g'(x)\ln{\left[f(x)\right]}\right\}

    And since h(x) = f(x)^{g(x)} this means

    h'(x) = f(x)^{g(x)}\left\{\frac{g(x)f'(x)}{f(x)} + g'(x)\ln{\left[f(x)\right]}\right\}.


    e) If h(x) = x^x

    Let f(x) = x so that f'(x) = 1.

    Let g(x) = x so that g'(x) = 1.


    h'(x) = f(x)^{g(x)}\left\{\frac{g(x)f'(x)}{f(x)} + g'(x)\ln{\left[f(x)\right]}\right\}

     = x^x\left[\frac{x\cdot 1}{x} + 1\cdot \ln{x}\right]

     = x^x\left(1 + \ln{x}\right).
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