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Math Help - Differential equation

  1. #1
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    Differential equation

    Given y = f(x) prove that the derivative of y^2 with respect to x^2 is given by dy/dx.y/x .Hence find the derivative of y^2 with respect to x^2 given f(x) = [x^2][e^(-x^3)] - xe^(-x)
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  2. #2
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    Use the chain rule. Let u= x^2. Then \frac{df}{dx}= \frac{df}{du}\frac{du}{dx}= 2x\frac{df}{du} so \frac{df}{d(x^2)}= \frac{df}{du}= \frac{df}{dx}\frac{1}{2x}.
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  3. #3
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    er .. u misunderstanding the question .?? >> Prove that the derivative of y^2with respect to x^2is given by \frac{dy}{dx}.\frac{y}{x} .
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  4. #4
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    This is my solution ... anyone can confirm for me ..??
    y = f(x)

    x = g(y) where g is inverse function of f

    \frac{dy}{dx} = f'(x)

    \frac{dx}{dy} = g'(y)

    y^2 = [f(x)]^2

    x^2 = [g(y)]^2

    \frac{dy^2}{dx} = 2[f(x)][f'(x)]

    \frac{dx^2}{dy} = 2[g(y)][g'(y)]

    \frac{\frac{dy^2}{dx}}{\frac{dx^2}{dy}} = \frac{2[f(x)][f'(x)]}{2[g(y)][g'(y)]}

    \frac{dy^2}{dx^2} . \frac{dy}{dx} = \frac{y}{x} . \frac{\frac{dy}{dx}}{\frac{dx}{dy}}

    \frac{dy^2}{dx^2} . \frac{dy}{dx} = \frac{y}{x} . \frac{dy}{dx} . \frac{dy}{dx}

    \frac{dy^2}{dx^2} = \frac{y}{x} . \frac{dy}{dx}(proven)

    f(x) = x^2e^{-X^3} - xe^{-x}

    f'(x) = e^{-x^3}(2x-3x^2) + e^{-x}(x-1)

    \frac{dy^2}{dx^2} = \frac{y}{x}[e^{-x^3}(2x-3x^2) + e^{-x}(x-1)]

    can it be this way ..?? last few step not sure ..
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