Second Order Derivatives

Concept of Second Order Derivative

Let $y = f(x)$ be a differentiable function. Its first derivative with respect to $x$ is $$\frac{dy}{dx} = f'(x).$$ If $f'(x)$ is itself differentiable, then we can differentiate once again with respect to $x$. This gives the second order derivative of $y$ with respect to $x$.

Notation: The second order derivative is written in any of the following forms:

• $\frac{d^2y}{dx^2}$
• $f''(x)$
• $y''$
• $D^2y$

Mathematically, $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right).$$ It measures the rate of change of the first derivative.

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Second Order Derivative of Parametric Functions

A common mistake is to think that for parametric equations $x=f(t)$ and $y=g(t)$, $$\frac{d^2y}{dx^2} = \frac{g''(t)}{f''(t)}.$$ This is not correct.

To find the second derivative correctly:

1. First compute $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$ Suppose this becomes some function of $t$, say $h(t)$.
2. Then differentiate this with respect to $x$ using the chain rule: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}[h(t)] \cdot \frac{dt}{dx}.$$ Since $$\frac{dt}{dx} = \frac{1}{dx/dt},$$ we get $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}.$$

Always remember: after differentiating with respect to $t$, divide by $\frac{dx}{dt}$.

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3. Proving Differential Equations

Many problems in this topic ask you to prove an identity involving $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$. Such relations are called differential equations.

Useful strategies

• Direct substitution: Find $y'$ and $y''$, then substitute them into the required expression and simplify.
• Smart manipulation: Sometimes it is better to first rearrange the first derivative into a simpler form and then differentiate again. This often avoids complicated quotient-rule work.

In board and entrance exam problems, both methods are useful depending on the form of the function.

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Solved Examples

Example 1: Basic Second Order Derivative

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

Solution: Step 1: Differentiate once. $$\frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(\tan x) = 3x^2 + \sec^2 x.$$

Step 2: Differentiate again. $$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2) + \frac{d}{dx}(\sec^2 x) = 6x + \frac{d}{dx}(\sec^2 x).$$

Step 3: Use the chain rule on $\sec^2 x$. $$\frac{d}{dx}(\sec^2 x) = 2\sec x\cdot \frac{d}{dx}(\sec x) = 2\sec x(\sec x\tan x)=2\sec^2 x\tan x.$$

Step 4: Therefore, $$\frac{d^2y}{dx^2} = 6x + 2\sec^2 x\tan x.$$

Answer: $6x + 2\sec^2 x\tan x$

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Example 2: Product Rule

Find the second order derivative of $y = x^2 e^x$.

Solution: Step 1: Differentiate once using the product rule. $$\frac{dy}{dx} = x^2\frac{d}{dx}(e^x) + e^x\frac{d}{dx}(x^2) = x^2 e^x + 2x e^x = e^x(x^2+2x).$$

Step 2: Differentiate again. $$\frac{d^2y}{dx^2} = \frac{d}{dx}\big[e^x(x^2+2x)\big].$$ Applying the product rule again, $$\frac{d^2y}{dx^2} = e^x(x^2+2x) + e^x(2x+2).$$

Step 3: Factor out $e^x$. $$\frac{d^2y}{dx^2} = e^x\big(x^2+2x+2x+2\big)=e^x(x^2+4x+2).$$

Answer: $e^x(x^2 + 4x + 2)$

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Example 3: Logarithmic Function

Find $\frac{d^2y}{dx^2}$ for $y = \log(\log x)$.

Solution: Step 1: Differentiate once using the chain rule. $$\frac{dy}{dx} = \frac{1}{\log x}\cdot \frac{d}{dx}(\log x)=\frac{1}{\log x}\cdot \frac{1}{x}=\frac{1}{x\log x}.$$

Step 2: Write the first derivative as $$\frac{dy}{dx}=(x\log x)^{-1}.$$ Then differentiate again: $$\frac{d^2y}{dx^2} = - (x\log x)^{-2}\cdot \frac{d}{dx}(x\log x).$$

Step 3: Differentiate $x\log x$ using the product rule. $$\frac{d}{dx}(x\log x)=x\cdot \frac{1}{x}+\log x \cdot 1 = 1+\log x.$$

Step 4: Substitute back. $$\frac{d^2y}{dx^2} = -\frac{1+\log x}{(x\log x)^2}.$$

Answer: $-\frac{1 + \log x}{(x \log x)^2}$

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Example 4: Proving a Differential Equation

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

Solution: Step 1: Differentiate once. $$\frac{dy}{dx} = A\cos x - B\sin x.$$

Step 2: Differentiate again. $$\frac{d^2y}{dx^2} = -A\sin x - B\cos x.$$

Step 3: Add $y$ to $y''$. $$\frac{d^2y}{dx^2} + y = (-A\sin x - B\cos x) + (A\sin x + B\cos x)=0.$$

Hence proved.

Answer: Proved.

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Example 5: Proving with Exponential Functions

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

Solution: Step 1: Differentiate once. $$\frac{dy}{dx}=3(2e^{2x})+2(3e^{3x})=6e^{2x}+6e^{3x}.$$

Step 2: Differentiate again. $$\frac{d^2y}{dx^2}=12e^{2x}+18e^{3x}.$$

Step 3: Substitute into the given expression. $$\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y$$ $$=(12e^{2x}+18e^{3x})-5(6e^{2x}+6e^{3x})+6(3e^{2x}+2e^{3x}).$$

Step 4: Simplify the coefficients. $$= e^{2x}(12-30+18)+e^{3x}(18-30+12)=0+0=0.$$

Hence proved.

Answer: Proved.

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Example 6: Finding Second Derivative in Terms of $y$ Alone

If $y = \cos^{-1} x$, find $\frac{d^2y}{dx^2}$ in terms of $y$ alone.

Solution: Step 1: From $$y = \cos^{-1}x,$$ we get $$x = \cos y.$$

Step 2: Differentiate with respect to $y$. $$\frac{dx}{dy} = -\sin y.$$ Hence, $$\frac{dy}{dx} = \frac{1}{dx/dy} = -\frac{1}{\sin y} = -\csc y.$$

Step 3: Differentiate again with respect to $x$. Since the expression is in terms of $y$, use the chain rule: $$\frac{d^2y}{dx^2} = \frac{d}{dx}(-\csc y)=\frac{d}{dy}(-\csc y)\cdot \frac{dy}{dx}.$$ Now, $$\frac{d}{dy}(-\csc y)=\csc y\cot y.$$ So, $$\frac{d^2y}{dx^2}=\csc y\cot y\cdot (-\csc y)=-\csc^2 y\cot y.$$

Answer: $-\csc^2 y \cot y$

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Example 7: Second Derivative of Parametric Equations

Find $\frac{d^2y}{dx^2}$, if $x = at^2$ and $y = 2at$.

Solution: Step 1: Differentiate with respect to $t$. $$\frac{dx}{dt}=2at, \qquad \frac{dy}{dt}=2a.$$

Step 2: Find the first derivative. $$\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}=t^{-1}.$$

Step 3: Differentiate $\frac{dy}{dx}$ with respect to $t$. $$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dt}(t^{-1})=-t^{-2}.$$

Step 4: Divide by $dx/dt$ to get the second derivative. $$\frac{d^2y}{dx^2}=\frac{-t^{-2}}{2at}=-\frac{1}{2at^3}.$$

Answer: $-\frac{1}{2at^3}$

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Example 8: Inverse Trigonometric Differential Equation Proof

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

Solution: Step 1: Differentiate once. $$\frac{dy}{dx} = 2\sin^{-1}x \cdot \frac{1}{\sqrt{1-x^2}}.$$ Let $y_1 = \frac{dy}{dx}$. Then, $$\sqrt{1-x^2}\, y_1 = 2\sin^{-1}x.$$

Step 2: Square both sides. $$(1-x^2)y_1^2 = 4(\sin^{-1}x)^2 = 4y.$$

Step 3: Differentiate implicitly with respect to $x$. $$\frac{d}{dx}\big[(1-x^2)y_1^2\big] = \frac{d}{dx}(4y).$$ Using the product rule, $$(-2x)y_1^2 + (1-x^2)\cdot 2y_1 y_2 = 4y_1,$$ where $y_2 = \frac{d^2y}{dx^2}$.

Step 4: Divide throughout by $2y_1$. $$-xy_1 + (1-x^2)y_2 = 2.$$

Step 5: Rearrange. $$(1-x^2)y_2 - xy_1 - 2 = 0.$$ Thus, $$(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - 2 = 0.$$

Answer: Proved.

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Example 9: Second Order Derivative of Implicit Function

If $x^2 + y^2 = a^2$, find $\frac{d^2y}{dx^2}$.

Solution: Step 1: Differentiate implicitly. $$2x + 2y\frac{dy}{dx}=0$$ so $$\frac{dy}{dx} = -\frac{x}{y}.$$

Step 2: Differentiate again. $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right).$$ Using the quotient rule, $$\frac{d^2y}{dx^2} = -\left(\frac{y\cdot 1 - x\cdot y'}{y^2}\right).$$

Step 3: Substitute $y' = -\frac{x}{y}$. $$\frac{d^2y}{dx^2} = -\left(\frac{y - x\left(-\frac{x}{y}\right)}{y^2}\right)= -\left(\frac{y + \frac{x^2}{y}}{y^2}\right).$$

Step 4: Simplify. $$\frac{d^2y}{dx^2} = -\frac{x^2+y^2}{y^3}.$$ Since $x^2+y^2=a^2$, $$\frac{d^2y}{dx^2} = -\frac{a^2}{y^3}.$$

Answer: $-\frac{a^2}{y^3}$

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Example 10: Parametric Trigonometric Second Derivative

Find $\frac{d^2y}{dx^2}$, if $x = a\cos\theta$ and $y = b\sin\theta$.

Solution: Step 1: Differentiate with respect to $\theta$. $$\frac{dx}{d\theta} = -a\sin\theta, \qquad \frac{dy}{d\theta} = b\cos\theta.$$

Step 2: Find the first derivative. $$\frac{dy}{dx} = \frac{b\cos\theta}{-a\sin\theta} = -\frac{b}{a}\cot\theta.$$

Step 3: Differentiate $\frac{dy}{dx}$ with respect to $\theta$. $$\frac{d}{d\theta}\left(-\frac{b}{a}\cot\theta\right)= -\frac{b}{a}(-\csc^2\theta)=\frac{b}{a}\csc^2\theta.$$

Step 4: Divide by $dx/d\theta$. $$\frac{d^2y}{dx^2} = \frac{\frac{b}{a}\csc^2\theta}{-a\sin\theta} = -\frac{b}{a^2}\cdot \frac{1}{\sin^3\theta} = -\frac{b}{a^2}\csc^3\theta.$$

Answer: $-\frac{b}{a^2}\csc^3\theta$