CBSE Sample Paper 2024-25 · Section B · 2 marks

Q23: Derivative of $\tan^{-1}x$ with respect to $\log x$ ⁠, or of $(\cos x)^x$

Sample 2024-25Ch 5 · Continuity and Differentiability

The question

(a) Find the derivative of $\tan^{-1}x$ with respect to $\log x$ ⁠; (where $x \in (1, \infty)$ ⁠).

(b) Differentiate the following function with respect to x : $(\cos x)^x$ ⁠; $\left(\text{where } x \in \left(0, \dfrac{\pi}{2}\right)\right)$ ⁠.

Exam answer

(a)

Let $y = \tan^{-1}x$ and $z = \log x$ ⁠.

Why?

To differentiate one function with respect to another, give each one a name. Then differentiate both with respect to the common variable $x$ ⁠.

\text{Then } \frac{dy}{dx} = \frac{1}{1 + x^2}

\text{and } \frac{dz}{dx} = \frac1x

Why?

These are standard derivatives:

\begin{aligned} \frac{d}{dx}\tan^{-1}x &= \frac{1}{1 + x^2} \\ \frac{d}{dx}\log x &= \frac1x \end{aligned}

\begin{aligned} \text{So, } \frac{dy}{dz} &= \frac{\dfrac{dy}{dx}}{\dfrac{dz}{dx}} = \frac{\dfrac{1}{1 + x^2}}{\dfrac1x} \\ &= \frac{x}{1 + x^2} \end{aligned}

Why?

By the chain rule,

\frac{dy}{dx} = \frac{dy}{dz}\cdot\frac{dz}{dx}

Dividing both sides by $\frac{dz}{dx}$ gives the formula used. This is allowed because $\frac1x \ne 0$ ⁠.

Dividing by $\frac1x$ is the same as multiplying by $x$ ⁠.

\begin{aligned} \frac{1}{1 + x^2} \div \frac1x &= \frac{1}{1 + x^2} \times \frac{x}{1} \\ &= \frac{x}{1 + x^2} \end{aligned}

The derivative of $\tan^{-1}x$ w.r.t. $\log x$ is $\dfrac{x}{1 + x^2}$ ⁠.

(b)

Let $y = (\cos x)^x$ ⁠. Then $y = e^{x\log\cos x}$ ⁠.

Why?

Here both the base $\cos x$ and the power $x$ change with $x$ ⁠. Neither the power rule nor the rule for $a^x$ applies, so we first rewrite the function with base $e$ ⁠.

For $a > 0$ ⁠, $a = e^{\log a}$ ⁠, so

a^b = e^{b\log a}

On $\left(0, \frac{\pi}{2}\right)$ ⁠, $\cos x > 0$ ⁠, so $\log\cos x$ is defined.

Differentiating both sides w.r.t. $x$ ⁠,

\frac{dy}{dx} = e^{x\log\cos x}\frac{d}{dx}(x\log\cos x)

Why?

Chain rule for the exponential:

\frac{d}{dx}e^u = e^u\frac{du}{dx}

with $u = x\log\cos x$ ⁠.

\begin{aligned} &\Rightarrow \frac{dy}{dx} \\ &= (\cos x)^x\bigg\{\log\cos x\cdot\frac{d}{dx}(x) \\ &\quad + x\frac{d}{dx}(\log\cos x)\bigg\} \\ &= (\cos x)^x\bigg\{\log\cos x \\ &\quad + x\cdot\frac{1}{\cos x}(-\sin x)\bigg\} \end{aligned}

Why?

$e^{x\log\cos x}$ is $(\cos x)^x$ again. The bracket is a product, so use the product rule:

\frac{d}{dx}(uv) = \frac{du}{dx}v + u\frac{dv}{dx}

$\frac{d}{dx}(x) = 1$ ⁠, and by the chain rule

\begin{aligned} &\frac{d}{dx}(\log\cos x) \\ &= \frac{1}{\cos x}\cdot(-\sin x) \end{aligned}

With $u = x$ ⁠, $v = \log\cos x$ ⁠, $u' = 1$ ⁠, $v' = \frac{1}{\cos x}(-\sin x)$ ⁠:

\begin{aligned} &\frac{d}{dx}(x\log\cos x) \\ &= (1)\log\cos x + x\cdot\frac{-\sin x}{\cos x} \end{aligned}

Finally $\frac{\sin x}{\cos x} = \tan x$ ⁠, so the second term is $-x\tan x$ ⁠.

\begin{aligned} x\cdot\frac{1}{\cos x}(-\sin x) &= -x\cdot\frac{\sin x}{\cos x} \\ &= -x\tan x \end{aligned}

$\therefore\ \dfrac{dy}{dx} = (\cos x)^x$ $(\log\cos x - x\tan x)$ ⁠.

Where this comes from

Chapter 5, Continuity and Differentiability: Derivative w.r.t. a function; logarithmic form.

Question type: (a) Parametric form; (b) Logarithmic differentiation

How to spot it: click to see

(a) Type 8. Parametric form: $x$ and $y$ given through $t$ or $\theta$

Spot it: two equations, one for $x$ and one for $y$ ⁠, both in the same third letter.

Here: Derivative of $\tan^{-1}x$ with respect to $\log x$ ⁠: both are given through $x$ ⁠, so $x$ plays the parameter. Divide $\frac{1}{1 + x^2}$ by $\frac1x$ to get $\frac{x}{1 + x^2}$ ⁠.

(b) Type 7. Logarithmic differentiation: $x$ in the power

Spot it: a power whose exponent contains $x$ ⁠, like $x^x$ ⁠, $x^{\sin x}$ or $(\log x)^{\cos x}$ ⁠, or a long product.

Here: $x$ is the power of $\cos x$ ⁠: write $(\cos x)^x = e^{x\log\cos x}$ ⁠, which is the same as taking log, and differentiate $x\log\cos x$ by the product rule.

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Q23: Derivative of $\tan^{-1}x$ with respect to $\log x$ ⁠, or of $(\cos x)^x$