Question

\(y=x^{2}\) \(\frac{d y}{d x}=2 x \quad \frac{d x}{d y}=\frac{1}{2 x}\) $\frac{d^{2} y}{d x^{2}}=2 \quad \frac{d^{2} x}{d y^{2}}=\frac{-1 \times 2}{(2 x)^{2}} \times \frac{d x}{d y}=\frac{-1}{2 x^{2}} \times \frac{1}{2 x}$ \(\frac{d^{2} y}{d x^{2}} \cdot \frac{d^{2} x}{d y^{2}}=\frac{-1}{2 x^{3}}\)

Short answer

Question: Verify that the product of the second derivatives of y with respect to x and x with respect to y is equal to \(\frac{-1}{2x^3}\) for the function \(y = x^2\).

Step-by-step solution

Find the first derivative of y with respect to x

We are given the first derivative of y with respect to x as \(\frac{dy}{dx} = 2x\).

Find the second derivative of y with respect to x

We are given the second derivative of y with respect to x as \(\frac{d^2y}{dx^2} = 2\).

Find the first derivative of x with respect to y

We are given the first derivative of x with respect to y as \(\frac{dx}{dy} = \frac{1}{2x}\).

Find the second derivative of x with respect to y

We are given the second derivative of x with respect to y as \(\frac{d^2x}{dy^2} = \frac{-1}{2x^{2}}\times\frac{1}{2x}\).

Find the product of the second derivatives

Multiply the second derivative of y with respect to x with the second derivative of x with respect to y: \(\frac{d^2y}{dx^2}\cdot\frac{d^2x}{dy^2} = (2) \cdot \left(\frac{-1}{2x^{2}}\times\frac{1}{2x}\right) = \frac{-1}{2x^3}\). The product of the second derivatives is verified to be equal to \(\frac{-1}{2x^3}\).