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Q13:
Find $\frac{dy}{dx}$ in the following: $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right), -1 < x < 1$
Solution :
Step 1: Substitution and Domain Analysis
Given the function $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)$, we observe the structure of the internal algebraic expression $\frac{2x}{1+x^2}$. This form suggests the trigonometric identity for $\sin(2\theta)$ when we substitute $x = \tan\theta$.
Let $x = \tan\theta$. Given $-1 < x < 1$, the corresponding range for $\theta$ is $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.
[By trigonometric substitution: $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$]
Step 2: Simplifying the Expression
Substitute $x = \tan\theta$ into the function:
$y = \cos^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)$ $y = \cos^{-1}(\sin(2\theta))$To evaluate $\cos^{-1}(\sin(2\theta))$, we convert sine to cosine using the co-function identity $\sin(\alpha) = \cos\left(\frac{\pi}{2} - \alpha\right)$:
$y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right)$Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, then $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$. Multiplying by $-1$, we get $-\frac{\pi}{2} < -2\theta < \frac{\pi}{2}$. Adding $\frac{\pi}{2}$, we obtain $0 < \frac{\pi}{2} - 2\theta < \pi$.
[Per the definition of the principal value branch of $\cos^{-1}(x)$, which is $[0, \pi]$, we can simplify the expression directly.]
$y = \frac{\pi}{2} - 2\theta$Step 3: Back-Substitution and Differentiation
Since $x = \tan\theta$, it follows that $\theta = \tan^{-1}(x)$. Substituting this back into the equation for $y$:
$y = \frac{\pi}{2} - 2\tan^{-1}(x)$Now, differentiate with respect to $x$ using the linearity property of the derivative:
$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - 2 \frac{d}{dx}\left(\tan^{-1}(x)\right)$[Per the Constant Rule, $\frac{d}{dx}(C) = 0$, and per the standard derivative rule, $\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}$]
$\frac{dy}{dx} = 0 - 2\left(\frac{1}{1+x^2}\right)$ $\frac{dy}{dx} = -\frac{2}{1+x^2}$Visual Representation of the Function Gradient
Final Solution: $\frac{dy}{dx} = -\frac{2}{1+x^2}$
More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.3
- Q1: Find $\frac{dy}{dx}$ in the following: $2x + 3y = \sin x$
- Q2: Find $\frac{dy}{dx}$ in the following: $2x + 3y = \sin y$
- Q3: Find $\frac{dy}{dx}$ in the following: $ax + by^2 = \cos y$
- Q4: Find $\frac{dy}{dx}$ in the following: $xy + y^2 = \tan x + y$
- Q5: Find $\frac{dy}{dx}$ in the following: $x^2 + xy + y^2 = 100$
- Q6: Find $\frac{dy}{dx}$ in the following: $x^3 + x^2y + xy^2 + y^3 = 81$
- Q7: Find $\frac{dy}{dx}$ in the following: $\sin^2 y + \cos xy = \kappa$
- Q8: Find $\frac{dy}{dx}$ in the following: $\sin^2 x + \cos^2 y = 1$
- Q9: Find $\frac{dy}{dx}$ in the following: $y = \sin^{-1} \left(\frac{2x}{1+x^2}\right)$
- Q10: Find $\frac{dy}{dx}$ in the following: $y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right), -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$
- Q11: Find $\frac{dy}{dx}$ in the following: $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$
- Q12: Find $\frac{dy}{dx}$ in the following: $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$
- Q14: Find $\frac{dy}{dx}$ in the following: $y = \sin^{-1}(2x\sqrt{1-x^2}), -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$
- Q15: Find $\frac{dy}{dx}$ in the following: $y = \sec^{-1}\left(\frac{1}{2x^2-1}\right), 0 < x < \frac{1}{\sqrt{2}}$
CBSE Solutions for Class 12 Mathematics Continuity and Differentiability
- Continuity and Differentiability Class 12 EXERCISE 5.1
- Continuity and Differentiability Class 12 EXERCISE 5.2
- Continuity and Differentiability Class 12 EXERCISE 5.4
- Continuity and Differentiability Class 12 EXERCISE 5.5
- Continuity and Differentiability Class 12 EXERCISE 5.6
- Continuity and Differentiability Class 12 EXERCISE 5.7
Chapters in CBSE - Class 12 Mathematics
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