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)
This is my solution ... anyone can confirm for me ..??
$\displaystyle y = f(x)$
$\displaystyle x = g(y) $where g is inverse function of f
$\displaystyle \frac{dy}{dx} = f'(x)$
$\displaystyle \frac{dx}{dy} = g'(y)$
$\displaystyle y^2 = [f(x)]^2$
$\displaystyle x^2 = [g(y)]^2$
$\displaystyle \frac{dy^2}{dx} = 2[f(x)][f'(x)]$
$\displaystyle \frac{dx^2}{dy} = 2[g(y)][g'(y)]$
$\displaystyle \frac{\frac{dy^2}{dx}}{\frac{dx^2}{dy}} = \frac{2[f(x)][f'(x)]}{2[g(y)][g'(y)]}$
$\displaystyle \frac{dy^2}{dx^2} . \frac{dy}{dx} = \frac{y}{x} . \frac{\frac{dy}{dx}}{\frac{dx}{dy}} $
$\displaystyle \frac{dy^2}{dx^2} . \frac{dy}{dx} = \frac{y}{x} . \frac{dy}{dx} . \frac{dy}{dx}$
$\displaystyle \frac{dy^2}{dx^2} = \frac{y}{x} . \frac{dy}{dx}$(proven)
$\displaystyle f(x) = x^2e^{-X^3} - xe^{-x}$
$\displaystyle f'(x) = e^{-x^3}(2x-3x^2) + e^{-x}(x-1)$
$\displaystyle \frac{dy^2}{dx^2} = \frac{y}{x}[e^{-x^3}(2x-3x^2) + e^{-x}(x-1)]$
can it be this way ..?? last few step not sure ..