Q63 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\text { If } y=\sin ^{-1}\left(\dfrac{5 x+12 \sqrt{1-x^2}}{13}\right) \text { then } \dfrac{d y}{d x} \text { equals }$
- A$\dfrac{-1}{\sqrt{1+x^2}}$
- B$\dfrac{-2 x}{\sqrt{1-x^2}}$
- Correct answerC$\dfrac{1}{\sqrt{1-x^2}}$
- D$\dfrac{2 x}{\sqrt{1-x^2}}$
Solution
Given $y = \sin^{-1} \left( \dfrac{5x + 12\sqrt{1-x^2}}{13} \right)$.
Let $x = \sin \theta$, where $\theta = \sin^{-1} x$. Then $\sqrt{1-x^2} = \cos \theta$.
Substituting these into the expression for $y$:
$y = \sin^{-1} \left( \dfrac{5 \sin \theta + 12 \cos \theta}{13} \right)$
Let $\cos \alpha = \dfrac{5}{13}$, which implies $\sin \alpha = \dfrac{12}{13}$.
$y = \sin^{-1} (\sin \theta \cos \alpha + \cos \theta \sin \alpha)$
$y = \sin^{-1} (\sin(\theta + \alpha))$
$y = \theta + \alpha = \sin^{-1} x + \cos^{-1} \left( \dfrac{5}{13} \right)$
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = \dfrac{d}{dx} (\sin^{-1} x) + \dfrac{d}{dx} (\cos^{-1} \dfrac{5}{13})$
Since $\cos^{-1} \dfrac{5}{13}$ is a constant, its derivative is $0$.
$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-x^2}}$
Answer: $\dfrac{1}{\sqrt{1-x^2}}$
Let $x = \sin \theta$, where $\theta = \sin^{-1} x$. Then $\sqrt{1-x^2} = \cos \theta$.
Substituting these into the expression for $y$:
$y = \sin^{-1} \left( \dfrac{5 \sin \theta + 12 \cos \theta}{13} \right)$
Let $\cos \alpha = \dfrac{5}{13}$, which implies $\sin \alpha = \dfrac{12}{13}$.
$y = \sin^{-1} (\sin \theta \cos \alpha + \cos \theta \sin \alpha)$
$y = \sin^{-1} (\sin(\theta + \alpha))$
$y = \theta + \alpha = \sin^{-1} x + \cos^{-1} \left( \dfrac{5}{13} \right)$
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = \dfrac{d}{dx} (\sin^{-1} x) + \dfrac{d}{dx} (\cos^{-1} \dfrac{5}{13})$
Since $\cos^{-1} \dfrac{5}{13}$ is a constant, its derivative is $0$.
$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-x^2}}$
Answer: $\dfrac{1}{\sqrt{1-x^2}}$
Hence, the correct answer is option: 'C'.