Second Order Derivative

Let y = f(x). If the first derivative \[\frac{dy}{dx} = f'(x)\] is itself differentiable, then differentiating once again with respect to \[x\] gives the second order derivative.

If \[y = \text{A} \sin x + \text{B} \cos x\], then prove that \[\frac{d^{2}y}{dx^{2}} + y = 0\].

Solution: We have

and

Hence \[\frac{d^{2}y}{dx^{2}} + y = 0\]

If \[y = \sin^{-1} x\], show that \[(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0\].

Solution: We have \[y = \sin^{-1} x\]. Then

or 

So

\[\frac{d}{dx} \left( \sqrt{(1 - x^2)} \cdot \frac{dy}{dx} \right) = 0\]

or

or

Hence \[(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0\]

Alternatively, Given that \[y = \sin^{-1} x\], we have

So\[(1 - x^2) \cdot 2y_1 y_2 + y_1^2 (0 - 2x) = 0\]

Hence \[(1 - x^2) y_2 - xy_1 = 0\]

• Second derivative means differentiating the function twice with respect to the same variable.

• It is defined only when the first derivative is differentiable.

• Common notations are \[\frac{d^2y}{dx^2}\], f''(x), y'', \[D^2y\], and \[y_2\].

• Higher order derivatives can be defined similarly.