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 $\displaystyle u= x^2$. Then $\displaystyle \frac{df}{dx}= \frac{df}{du}\frac{du}{dx}= 2x\frac{df}{du}$ so $\displaystyle \frac{df}{d(x^2)}= \frac{df}{du}= \frac{df}{dx}\frac{1}{2x}$.

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

4. 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 ..