NCERT Class 12 Mathematics — Continuity and Differentiability, Exercise 5.7. All 17 questions solved.

The second order derivative is nothing new — it is the derivative of the derivative:

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

also written $y”$, $y_2$ or $\mathrm{D}^2y$.

The exercise splits into ten straight computations and seven “show that” problems where a given function is shown to satisfy a differential equation.

Key insight. For the proof questions, do not substitute the explicit derivatives into the equation and grind. Instead, eliminate the awkward parts using the original relation.

Question 13 is the model: from $y = 3\cos(\log x) + 4\sin(\log x)$ you get $xy_1 = -3\sin(\log x) + 4\cos(\log x)$, and differentiating that gives $x^2y_2 + xy_1 = -y$ directly, because the bracket that appears is the original $y$ again. Aim for that structure and the algebra stays short.

Find the second order derivatives of the functions given in Exercises 1 to 10.

Question 1

$x^2 + 3x + 2$

Solution. $y_1 = 2x + 3$, so $y_2 = 2$.

Question 2

$x^{20}$

Solution. $y_1 = 20x^{19}$, so $y_2 = 380x^{18}$.

Question 3

$x \cdot \cos x$

Solution. By the product rule, $y_1 = \cos x – x\sin x$. Differentiating again, the second term needs the product rule too:

$$y_2 = -\sin x – \left(\sin x + x\cos x\right) = -x\cos x – 2\sin x$$

Question 4

$\log x$

Solution. $y_1 = \tfrac1x = x^{-1}$, so $y_2 = -x^{-2}$.

Question 5

$x^3\log x$

Solution. $y_1 = 3x^2\log x + x^2 = x^2(3\log x + 1)$. Differentiating again:

$$y_2 = 2x(3\log x + 1) + x^2 \cdot \frac3x = 6x\log x + 2x + 3x$$

Question 6

$e^x\sin 5x$

Solution. $y_1 = e^x\sin 5x + 5e^x\cos 5x = e^x(\sin 5x + 5\cos 5x)$.

$$y_2 = e^x(\sin 5x + 5\cos 5x) + e^x(5\cos 5x – 25\sin 5x) = e^x(10\cos 5x – 24\sin 5x)$$

Question 7

$e^{6x}\cos 3x$

Solution. $y_1 = e^{6x}(6\cos 3x – 3\sin 3x)$.

$$y_2 = 6e^{6x}(6\cos 3x – 3\sin 3x) + e^{6x}(-18\sin 3x – 9\cos 3x) = e^{6x}(27\cos 3x – 36\sin 3x)$$

Question 8

$\tan^{-1}x$

Solution. $y_1 = \dfrac{1}{1+x^2} = \left(1+x^2\right)^{-1}$, so by the chain rule

$$y_2 = -\left(1+x^2\right)^{-2} \cdot 2x$$

Question 9

$\log(\log x)$

Solution. $y_1 = \dfrac{1}{x\log x}$, which is a quotient:

$$y_2 = -\frac{\frac{d}{dx}(x\log x)}{(x\log x)^2} = -\frac{\log x + 1}{(x\log x)^2}$$

Question 10

$\sin(\log x)$

Solution. $y_1 = \dfrac{\cos(\log x)}{x}$, another quotient:

$$y_2 = \frac{-\frac{\sin(\log x)}{x} \cdot x – \cos(\log x)}{x^2}$$

Question 11

If $y = 5\cos x – 3\sin x$, prove that $\dfrac{d^2y}{dx^2} + y = 0$.

Solution. $y_1 = -5\sin x – 3\cos x$, and

$$y_2 = -5\cos x + 3\sin x = -(5\cos x – 3\sin x) = -y$$

Hence $y_2 + y = 0$. Every combination of $\sin x$ and $\cos x$ satisfies this equation, which is why it recurs throughout the differential equations chapter.

Question 12

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

Solution. $y_1 = \dfrac{-1}{\sqrt{1-x^2}}$, so

$$y_2 = \frac{-x}{\left(1-x^2\right)^{3/2}}$$

Now convert to $y$. Since $x = \cos y$ and $\sqrt{1-x^2} = \sin y$ (positive on $\cos^{-1}$’s branch $[0, \pi]$):

$$y_2 = \frac{-\cos y}{\sin^3y} = -\cot y\operatorname{cosec}^2y$$

Question 13

If $y = 3\cos(\log x) + 4\sin(\log x)$, show that $x^2y_2 + xy_1 + y = 0$.

Solution.

$$y_1 = \frac{-3\sin(\log x) + 4\cos(\log x)}{x} \quad\Longrightarrow\quad xy_1 = -3\sin(\log x) + 4\cos(\log x)$$

Multiplying by $x$ first is the move that keeps everything tidy. Differentiating this equation:

$$y_1 + xy_2 = \frac{-3\cos(\log x) – 4\sin(\log x)}{x} = \frac{-y}{x}$$

Multiplying through by $x$:

$$xy_1 + x^2y_2 = -y$$

which rearranges to the required result.

Question 14

If $y = \mathrm{A}e^{mx} + \mathrm{B}e^{nx}$, show that $\dfrac{d^2y}{dx^2} – (m+n)\dfrac{dy}{dx} + mny = 0$.

Solution.

$$y_1 = m\mathrm{A}e^{mx} + n\mathrm{B}e^{nx}, \qquad y_2 = m^2\mathrm{A}e^{mx} + n^2\mathrm{B}e^{nx}$$

Substituting into the left-hand side and collecting the $\mathrm{A}e^{mx}$ and $\mathrm{B}e^{nx}$ terms:

$$\mathrm{A}e^{mx}\left(m^2 – (m+n)m + mn\right) + \mathrm{B}e^{nx}\left(n^2 – (m+n)n + mn\right)$$

Each bracket is $m^2 – m^2 – mn + mn = 0$ and $n^2 – mn – n^2 + mn = 0$. Hence the whole expression vanishes.

The brackets are the quadratic $t^2 – (m+n)t + mn = (t-m)(t-n)$ evaluated at $t = m$ and $t = n$ — which is exactly why $m$ and $n$ were chosen as the exponents.

Question 15

If $y = 500e^{7x} + 600e^{-7x}$, show that $\dfrac{d^2y}{dx^2} = 49y$.

Solution. Each differentiation brings down a factor of $\pm7$, and squaring either gives $49$:

$$y_2 = 49 \cdot 500e^{7x} + 49 \cdot 600e^{-7x} = 49y$$

This is question 14 with $m = 7$ and $n = -7$, where $m + n = 0$ and $mn = -49$.

Question 16

If $e^y(x+1) = 1$, show that $\dfrac{d^2y}{dx^2} = \left(\dfrac{dy}{dx}\right)^2$.

Solution. Take logarithms first — the equation becomes much easier:

$$y + \log(x+1) = 0 \quad\Longrightarrow\quad y = -\log(x+1)$$

$$y_1 = \frac{-1}{x+1}, \qquad y_2 = \frac{1}{(x+1)^2}$$

And $\left(y_1\right)^2 = \dfrac{1}{(x+1)^2}$, which is $y_2$.

Question 17

If $y = \left(\tan^{-1}x\right)^2$, show that $\left(x^2+1\right)^2y_2 + 2x\left(x^2+1\right)y_1 = 2$.

Solution.

$$y_1 = 2\tan^{-1}x \cdot \frac{1}{1+x^2} \quad\Longrightarrow\quad \left(1+x^2\right)y_1 = 2\tan^{-1}x$$

Again, clear the denominator before differentiating again:

$$2xy_1 + \left(1+x^2\right)y_2 = \frac{2}{1+x^2}$$

Multiplying throughout by $\left(1+x^2\right)$:

$$2x\left(1+x^2\right)y_1 + \left(1+x^2\right)^2y_2 = 2$$

Common mistakes

• Differentiating only the first factor the second time. In question 3 the term $-x\sin x$ needs the product rule again, giving two terms, not one.
• Substituting explicit derivatives into a “show that”. Questions 13 and 17 are much shorter if you multiply out the denominator first and differentiate the resulting equation.
• Forgetting that $y_2$ requires the quotient or chain rule too. Questions 8, 9 and 10 all have a first derivative that is itself a quotient.
• Not converting fully to $y$. Question 12 asks for the answer “in terms of $y$ alone”, so every $x$ must be replaced using $x = \cos y$.
• Dropping a sign on $e^{-7x}$. In question 15 the first derivative has a $-7$; only after squaring does the sign disappear.
• Taking $\sqrt{1-x^2} = -\sin y$. On $\cos^{-1}$’s branch $[0, \pi]$ the sine is non-negative, so the positive root is correct.

Practise next

• Miscellaneous Exercise on Chapter 5 — questions 17, 18 and 22 there are second-derivative problems of the same kind.
• Exercise 9.2 — verifying that a function satisfies a differential equation, which is what questions 13 to 17 here are doing.