The Second Derivative

If $y = f(x)$ is differentiable and its derivative $f'(x)$ is itself differentiable, differentiating again gives the second order derivative:

$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$$

also written $y''$, $y_2$, or $f''(x)$. Nothing new is needed to compute it — every tool of the chapter (product rule, chain rule, implicit, parametric) simply gets applied a second time.

Quick specimens: $y = x^2 + 3x + 2$ gives $y' = 2x + 3$ and $y'' = 2$; $y = x^{20}$ gives $y'' = 20 \cdot 19\,x^{18} = 380x^{18}$; $y = \log x$ gives $y' = \frac{1}{x}$ and $y'' = -\frac{1}{x^2}$.

The exam staple: proving differential relations

Board papers rarely ask for $y''$ alone — they ask you to prove that $y$ satisfies a relation involving $y$, $y'$ and $y''$. The recipe: compute $y'$ and $y''$, substitute into the left side, and watch everything cancel.

1. $y = A\sin x + B\cos x$: here $y'' = -A\sin x - B\cos x = -y$, so $$y'' + y = 0$$
2. $y = 3e^{2x} + 2e^{3x}$: each exponential $e^{kx}$ contributes $k^2$ on double differentiation, and the combination satisfies $$y'' - 5y' + 6y = 0$$ (the numbers $2$ and $3$ are exactly the roots of $k^2 - 5k + 6 = 0$ — a useful pattern-check).
3. $y = \sin^{-1} x$: from $y' = \frac{1}{\sqrt{1 - x^2}}$, one more differentiation and a clean-up gives $$(1 - x^2)y'' - xy' = 0$$

A powerful shortcut for relation proofs: after the first differentiation, clear the radicals or denominators, then differentiate the cleaned equation implicitly. For $y = \sin^{-1} x$: write $\sqrt{1 - x^2}\,y' = 1$, differentiate both sides, and the target relation appears at once — no messy second quotient rule.

Parametric second derivatives — the trap

For $x = f(t)$, $y = g(t)$, the second derivative is not $\frac{d^2y/dt^2}{d^2x/dt^2}$. Differentiate $\frac{dy}{dx}$ (a function of $t$) with respect to $x$ via the chain rule:

$$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{1}{dx/dt}$$

— differentiate the slope w.r.t. the parameter, then divide by $\frac{dx}{dt}$ once more.

Solved Examples

Example 1 — A straight double differentiation

Find $\dfrac{d^2y}{dx^2}$ if $y = x^3 + \tan x$.

Step 1 — first derivative: $y' = 3x^2 + \sec^2 x$.

Step 2 — second derivative (chain rule on $\sec^2 x$): $y'' = 6x + 2\sec x \cdot \sec x\tan x$.

Answer: $y'' = 6x + 2\sec^2 x\tan x$.

Example 2 — The sine-cosine relation

If $y = A\sin x + B\cos x$, prove that $\dfrac{d^2y}{dx^2} + y = 0$.

Step 1 — differentiate twice: $y' = A\cos x - B\sin x$, then $y'' = -A\sin x - B\cos x$.

Step 2 — recognise: $y'' = -y$ exactly.

Answer: $y'' + y = 0$. ∎ Both constants vanish from the relation — that is the point: every combination $A\sin x + B\cos x$ satisfies the same equation.

Example 3 — The exponential relation

If $y = 3e^{2x} + 2e^{3x}$, prove that $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$.

Step 1 — differentiate twice: $y' = 6e^{2x} + 6e^{3x}$ and $y'' = 12e^{2x} + 18e^{3x}$.

Step 2 — substitute: $$y'' - 5y' + 6y = (12 - 30 + 18)e^{2x} + (18 - 30 + 12)e^{3x} = 0$$

Answer: proved. ∎ Pattern: $e^{kx}$ passes through $y'' - 5y' + 6y$ as the factor $k^2 - 5k + 6 = (k-2)(k-3)$, which vanishes precisely at $k = 2, 3$ — the exponents in $y$.

Example 4 — The inverse-sine relation via the clean-up trick

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

Step 1 — first derivative: $y' = \dfrac{1}{\sqrt{1 - x^2}}$.

Step 2 — clear the radical: $\sqrt{1 - x^2}\; y' = 1$.

Step 3 — differentiate the cleaned equation (product rule): $$\sqrt{1 - x^2}\;y'' + y'\cdot\frac{-x}{\sqrt{1 - x^2}} = 0$$

Step 4 — multiply through by $\sqrt{1 - x^2}$: $(1 - x^2)y'' - xy' = 0$. ∎

Answer: proved — the clean-up trick avoided differentiating a quotient with a radical, and delivered the relation in one line.

Example 5 — Second derivative of a product

Find $y''$ if $y = x\cos x$.

Step 1 — product rule once: $y' = \cos x - x\sin x$.

Step 2 — product rule again: $y'' = -\sin x - (\sin x + x\cos x) = -x\cos x - 2\sin x$.

Answer: $y'' = -(x\cos x + 2\sin x)$ — note the pattern: the original function reappears with a minus, plus a cross term.

Example 6 — Parametric second derivative done right

If $x = a(\cos t + t\sin t)$ and $y = a(\sin t - t\cos t)$, find $\dfrac{d^2y}{dx^2}$.

Step 1 — first derivatives w.r.t. $t$: $\frac{dx}{dt} = a\,t\cos t$ and $\frac{dy}{dt} = a\,t\sin t$ (the $\pm\sin t$, $\mp\cos t$ pairs cancel).

Step 2 — slope: $\frac{dy}{dx} = \tan t$.

Step 3 — differentiate the slope w.r.t. $t$, divide by $\frac{dx}{dt}$: $$\frac{d^2y}{dx^2} = \frac{\sec^2 t}{a\,t\cos t} = \frac{\sec^3 t}{a\,t}$$

Answer: $\dfrac{\sec^3 t}{at}$ — computed the honest way; dividing the second $t$-derivatives directly would give a wrong (and dimensionally senseless) answer.