CBSE Class 12 Mathematics Derivatives Q #1452
KNOWLEDGE BASED
REMEMBER
3 Marks 2025 AISSCE(Board Exam) SA
Differentiate $y=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ with respect to x, when $x\in(0,1)$.
Prev Next

AI Tutor Explanation

Powered by Gemini

Detailed Solution

Step 1: Substitution

Let $x = \tan\theta$. Since $x \in (0,1)$, we have $\theta \in (0, \frac{\pi}{4})$.

Step 2: Simplify the expression

Substitute $x = \tan\theta$ into the expression for $y$:\r\n$$y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)$$\r\nUsing the trigonometric identity $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$, we get:\r\n$$y = \cos^{-1}(\cos 2\theta)$$\r\nSince $\theta \in (0, \frac{\pi}{4})$, we have $2\theta \in (0, \frac{\pi}{2})$. Therefore, $\cos^{-1}(\cos 2\theta) = 2\theta$. So,\r\n$$y = 2\theta$$

Step 3: Express y in terms of x

Since $x = \tan\theta$, we have $\theta = \tan^{-1}x$. Therefore,\r\n$$y = 2\tan^{-1}x$$

Step 4: Differentiate y with respect to x

Differentiate $y$ with respect to $x$:\r\n$$\frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1}x) = 2\frac{d}{dx}(\tan^{-1}x)$$\r\nUsing the derivative of $\tan^{-1}x$, which is $\frac{1}{1+x^2}$, we get:\r\n$$\frac{dy}{dx} = 2\left(\frac{1}{1+x^2}\right) = \frac{2}{1+x^2}$$

Final Answer: $\frac{2}{1+x^2}$

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the derivative of inverse trigonometric functions and applying trigonometric identities.
Knowledge Dimension: PROCEDURAL
Justification: The question involves applying a series of steps, including trigonometric substitution, simplification using identities, and differentiation, to arrive at the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's ability to differentiate inverse trigonometric functions, a standard topic in the syllabus.

More from this Chapter

MCQ_SINGLE
Derivative of $\cos^{-1}(\frac{\sin x+\cos x}{\sqrt{2}})$, $-\frac{\pi}{4}
SUBJECTIVE
Amusement parks heavily rely on mathematical curves to design roller coaster tracks, balancing intense thrill with absolute safety. For a safe ride, the track must have no breaks (it must be continuous) and no sharp, sudden corners (it must be differentiable/smooth). An engineering team is designing a transition segment of a new coaster. The height $h(x)$ (in meters) of the track at a horizontal distance $x$ (in meters) from the starting platform is modeled by the following piecewise function: $$h(x) = \begin{cases} \frac{1}{2}x^2 + 2x, & 0 \leq x < 2 \\ ax + b, & 2 \leq x \leq 5 \end{cases}$$ Based on the given information, answer the following questions: (i) For the track to have no broken rails at the transition point $x = 2$, it must be continuous. Formulate the mathematical equation relating $a$ and $b$ to ensure this continuity. (1 Mark) (ii) Find the left-hand derivative (the slope of the track just before the transition point) at $x = 2$. (1 Mark) (iii) To ensure the transition at $x = 2$ is perfectly smooth (differentiable), calculate the exact values of $a$ and $b$ that the engineers must use. (2 Marks) — OR — (iii) If a junior engineer mistakenly designed the second track segment using $a = 3$ and $b = 0$, verify whether the track is mathematically safe (smooth and differentiable) at $x = 2$. Justify your answer. (2 Marks)
MCQ_SINGLE
The derivative of \(\tan^{-1}(x^{2})\) w.r.t. x is :
LA
For a positive constant 'a', differentiate $a^{t+\frac{1}{t}}$ with respect to $(t+\frac{1}{t})^{a}$ where t is a non-zero real number.
VSA
Differentiate $\left(\frac{5^{x}}{x^{5}}\right)$ with respect to x.
View All Questions