CBSE Sample Paper 2025-26 · Section B · 2 marks

Q22: If $y = \log\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$ ⁠, prove $\frac{dy}{dx} - \sec x = 0$

Sample 2025-26Ch 5 · Continuity and Differentiability

The question

If $y = \log\tan\left(\dfrac{\pi}{4} + \dfrac{x}{2}\right)$ ⁠, then prove that $\dfrac{dy}{dx} - \sec x = 0$

Exam answer

$y = \log\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$

Differentiating with respect to $x$ (chain rule),

\begin{aligned} \frac{dy}{dx} &= \frac{1}{\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)} \\ &\quad\cdot\sec^2\left(\frac{\pi}{4} + \frac{x}{2}\right)\cdot\frac12 \end{aligned}

Why?

$y$ is a function inside a function inside a function: log of tan of $\left(\frac{\pi}{4} + \frac{x}{2}\right)$ ⁠. The chain rule differentiates one layer at a time, from the outside in, and multiplies the results:

\begin{aligned} \frac{d}{dx}\log u &= \frac{1}{u}\cdot\frac{du}{dx} \\ \frac{d}{dx}\tan w &= \sec^2w\cdot\frac{dw}{dx} \end{aligned}

The innermost layer gives $\frac{d}{dx}\left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac12$ ⁠, since $\frac{\pi}{4}$ is a constant.

\begin{aligned} \frac{dy}{dx} &= \frac{\cos\left(\frac{\pi}{4} + \frac{x}{2}\right)}{\sin\left(\frac{\pi}{4} + \frac{x}{2}\right)} \\ &\quad\cdot\frac{1}{\cos^2\left(\frac{\pi}{4} + \frac{x}{2}\right)}\cdot\frac12 \\ &= \frac{1}{2\sin\left(\frac{\pi}{4} + \frac{x}{2}\right)\cos\left(\frac{\pi}{4} + \frac{x}{2}\right)} \end{aligned}

Why?

Write everything in sines and cosines:

\frac{1}{\tan A} = \frac{\cos A}{\sin A},\quad \sec^2A = \frac{1}{\cos^2A}

One $\cos A$ on the top cancels one of the two on the bottom, leaving $\sin A\cos A$ with the 2 from $\frac12$ ⁠.

With $A = \frac{\pi}{4} + \frac{x}{2}$ ⁠, multiply the three factors:

\begin{aligned} &\frac{\cos A}{\sin A}\cdot\frac{1}{\cos^2A}\cdot\frac12 \\ &= \frac{\cos A}{2\sin A\cos^2A} \\ &= \frac{\cos A}{2\sin A\cos A\cos A} \\ &= \frac{1}{2\sin A\cos A} \end{aligned}

Using $2\sin\theta\cos\theta = \sin 2\theta$ ⁠,

\begin{aligned} \frac{dy}{dx} &= \frac{1}{\sin\left(\frac{\pi}{2} + x\right)} \\ &= \frac{1}{\cos x} = \sec x \end{aligned}

Why?

Here $\theta = \frac{\pi}{4} + \frac{x}{2}$ ⁠, so $2\theta = \frac{\pi}{2} + x$ ⁠.

\begin{aligned} 2\theta &= 2\left(\frac{\pi}{4} + \frac{x}{2}\right) \\ &= \frac{2\pi}{4} + \frac{2x}{2} = \frac{\pi}{2} + x \end{aligned}

Then the allied-angle rule $\sin\left(\frac{\pi}{2} + x\right) = \cos x$ turns the denominator into $\cos x$ ⁠, and $\frac{1}{\cos x}$ is $\sec x$ by definition.

$\therefore\ \dfrac{dy}{dx} - \sec x = 0$ ⁠. Hence proved.

Where this comes from

Chapter 5, Continuity and Differentiability: Derivative of a composite function (chain rule).

Question type: The chain rule

How to spot it: click to see

Type 4. The chain rule: a function inside a function

Spot it: something other than a plain $x$ inside $\sin$ ⁠, $\cos$ ⁠, $\log$ ⁠, $e$ or a power, like $\sin(x^2 + 5)$ or $e^{\cos x}$ ⁠.

Here: Three layers: $\log$ of $\tan$ of $\frac\pi4 + \frac x2$ ⁠. Differentiate outside in, $\frac{1}{\tan(\ldots)}\cdot\sec^2(\ldots)\cdot\frac12$ ⁠, then use $2\sin\theta\cos\theta = \sin 2\theta$ to reach $\sec x$ ⁠.

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Continuity and Differentiability in past papers: question 20 of 35 · see them all

Q22: If y=log⁡tan⁡(π4+x2)y = \log\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)y=logtan(4π​+2x​)⁠, prove dydx−sec⁡x=0\frac{dy}{dx} - \sec x = 0dxdy​−secx=0

Q22: If $y = \log\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$ ⁠, prove $\frac{dy}{dx} - \sec x = 0$