NCERT Class 12 Mathematics — Continuity and Differentiability, Exercise 5.4. All 10 questions solved.

Two derivatives added to the toolkit, both used with the chain rule throughout:

$$\frac{d}{dx}e^x = e^x, \qquad \frac{d}{dx}\log x = \frac1x \quad (x > 0)$$

Key insight. Every answer in this exercise carries a domain restriction, and the restriction always comes from one of three places: a logarithm needs a positive argument, a square root needs a non-negative one, and a denominator must not vanish.

Question 8 is restricted to $x > 1$ because $\log(\log x)$ needs $\log x > 0$. Question 9 is restricted to $x > 0$ and implicitly to $x \ne 1$, since $\log x$ appears in the denominator. Writing the restriction is part of the answer, not an afterthought.

Differentiate the following w.r.t. $x$:

Question 1

$\dfrac{e^x}{\sin x}$

Solution. Quotient rule:

$$\frac{dy}{dx} = \frac{e^x\sin x – e^x\cos x}{\sin^2x}$$

The function is undefined wherever $\sin x = 0$.

Question 2

$e^{\sin^{-1}x}$

Solution. Chain rule with $\tfrac{d}{dx}\sin^{-1}x = \tfrac{1}{\sqrt{1-x^2}}$:

Question 3

$e^{x^3}$

Solution.

Question 4

$\sin\left(\tan^{-1}e^{-x}\right)$

Solution. Three layers: sine, arctangent, exponential.

$$\frac{dy}{dx} = \cos\left(\tan^{-1}e^{-x}\right) \cdot \frac{1}{1 + e^{-2x}} \cdot \left(-e^{-x}\right)$$

The middle factor is $\tfrac{1}{1+\left(e^{-x}\right)^2}$, which is where the $e^{-2x}$ comes from.

Question 5

$\log\left(\cos e^x\right)$

Solution.

$$\frac{dy}{dx} = \frac{1}{\cos e^x} \cdot \left(-\sin e^x\right) \cdot e^x = -e^x\tan e^x$$

The logarithm requires $\cos e^x > 0$, which excludes the points where $e^x$ is an odd multiple of $\tfrac{\pi}{2}$.

Question 6

$e^x + e^{x^2} + \cdots + e^{x^5}$

Solution. Differentiate term by term; the $k$-th term contributes $kx^{k-1}e^{x^k}$.

Question 7

$\sqrt{e^{\sqrt{x}}}$, $x > 0$

Solution. Write the function as $e^{\sqrt{x}/2}$ to make the chain shorter, or work through the root directly:

$$\frac{dy}{dx} = \frac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{4\sqrt{x}\sqrt{e^{\sqrt{x}}}}$$

Combining the two roots in the denominator gives $\sqrt{xe^{\sqrt{x}}}$.

Question 8

$\log(\log x)$, $x > 1$

Solution.

$$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac1x$$

The outer logarithm needs $\log x > 0$, which is exactly $x > 1$.

Question 9

$\dfrac{\cos x}{\log x}$, $x > 0$

Solution. Quotient rule:

$$\frac{dy}{dx} = \frac{-\sin x\log x – \cos x \cdot \frac1x}{(\log x)^2}$$

Multiplying numerator and denominator by $x$ to clear the inner fraction:

Question 10

$\cos\left(\log x + e^x\right)$, $x > 0$

Solution.

$$\frac{dy}{dx} = -\sin\left(\log x + e^x\right)\left(\frac1x + e^x\right)$$

Common mistakes

• Writing $\tfrac{d}{dx}e^{u} = e^{u}$ without the $u’$. Only $e^x$ itself differentiates to itself; every other exponent brings a factor.
• Omitting the domain restriction. Each of questions 1, 2, 5, 7, 8, 9 and 10 has one, and NCERT prints it as part of the answer.
• Getting $\tfrac{d}{dx}\log(\log x)$ wrong. It is $\tfrac{1}{\log x} \cdot \tfrac1x$, not $\tfrac{1}{x\log x} \cdot \tfrac1x$ or $\tfrac{1}{\log x}$.
• Mis-squaring the inner exponential. In question 4, $\left(e^{-x}\right)^2 = e^{-2x}$, not $e^{-x^2}$.
• Leaving a compound fraction unsimplified. Question 9’s answer only matches the key after multiplying through by $x$.
• Forgetting the minus sign from $e^{-x}$. Questions 4 and 5 both depend on it.

Practise next

• Exercise 5.5 — logarithmic differentiation, for functions with a variable in the exponent.
• Exercise 5.2 — the chain rule that every question here applies.