1. ## 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)

2. Use the chain rule. Let $u= x^2$. Then $\frac{df}{dx}= \frac{df}{du}\frac{du}{dx}= 2x\frac{df}{du}$ so $\frac{df}{d(x^2)}= \frac{df}{du}= \frac{df}{dx}\frac{1}{2x}$.

3. er .. u misunderstanding the question .?? >> Prove that the derivative of $y^2$with respect to $x^2$is given by $\frac{dy}{dx}.\frac{y}{x}$ .

4. 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 ..