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RavenCandelario
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Tyro
Asked: 2021-04-07T08:42:48+00:00 In: ,

Let f(x)=(c^x+d^x)^{1/x}, where c, d greater than 0. Find f'(x)

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Let f(x)=(c^x+d^x)^{1/x}, where c, d greater than 0. Find f'(x)

calculus

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  2. acedstud
    This answer was edited.

    Given: f(x)=(c^x+d^x)^{\frac{1}{x}}

    For simplicity sake, I will rewrite the function as y=(c^x+d^x)^{\frac{1}{x}}.

    To take care of the function in the exponent, we take the logarithm of both sides to get \ln{y}=\ln(c^x+d^x)^{\frac{1}{x}}=\dfrac{1}{x}\ln(c^x+d^x).

    Next, we differentiate both sides to get \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{x}\cdot\dfrac{1}{c^x+d^x}\cdot(c^x\ln{c}+d^x\ln{d})-\frac{1}{x^2}\ln(c^x+d^x)=\dfrac{c^x\ln{c}+d^x\ln{d}}{x(c^x+d^x)}-\dfrac{\ln(c^x+d^x)}{x^2}.

    Multiplying both sides by y gives \dfrac{dy}{dx}=y\left[\dfrac{c^x\ln{c}+d^x\ln{d}}{x(c^x+d^x)}-\dfrac{\ln(c^x+d^x)}{x^2}\right].

    But y=(c^x+d^x)^{\frac{1}{x}},

    therefore f'(x)=\dfrac{dy}{dx}=(c^x+d^x)^{\frac{1}{x}}\left(\dfrac{c^x\ln{c}+d^x\ln{d}}{x(c^x+d^x)}-\dfrac{\ln(c^x+d^x)}{x^2}\right).

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