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Q14:
Find $\frac{dy}{dx}$ in the following: $y = \sin^{-1}(2x\sqrt{1-x^2}), -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$
Solution :
Step 1: Trigonometric Substitution
To simplify the expression $y = \sin^{-1}(2x\sqrt{1-x^2})$, we use the trigonometric substitution method. Let $x = \sin\theta$. Consequently, $\theta = \sin^{-1}x$. [By Trigonometric Identity: $\sin^2\theta + \cos^2\theta = 1$].
Given the domain $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.
Step 2: Algebraic Simplification
Substitute $x = \sin\theta$ into the function:
$y = \sin^{-1}(2\sin\theta \sqrt{1-\sin^2\theta})$$y = \sin^{-1}(2\sin\theta \sqrt{\cos^2\theta})$
$y = \sin^{-1}(2\sin\theta \cos\theta)$
Utilizing the double-angle formula $\sin(2\theta) = 2\sin\theta\cos\theta$ [By Double Angle Identity]:
$y = \sin^{-1}(\sin 2\theta)$Step 3: Determining the Principal Value
Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, it follows that $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$.
Because the range of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we can simplify $\sin^{-1}(\sin 2\theta) = 2\theta$. [By definition of inverse trigonometric functions].
$y = 2\theta = 2\sin^{-1}x$Step 4: Differentiation
Now, differentiate $y$ with respect to $x$ using the standard derivative formula for inverse sine: $\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$. [By Derivative of Inverse Trigonometric Functions]:
$\frac{dy}{dx} = \frac{d}{dx}(2\sin^{-1}x)$$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\sin^{-1}x)$
$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$
Visual Representation of Substitution
Final Solution: \frac{dy}{dx} = \frac{2}{\sqrt{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: $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$
- Q13: Find $\frac{dy}{dx}$ in the following: $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right), -1 < x < 1$
- 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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