Limits, Continuity and Differentiability
80272
If \(x=\sin ^{-1}(\cos \theta)\) and \(y=\tan ^{-1} \theta\), then \(\frac{d y}{d x}=\)
1 \(\frac{1}{1+\theta^{2}}\)
2 \(1+\theta^{2}\)
3 \(\frac{-1}{1+\theta^{2}}\)
4 \(-\left(1+\theta^{2}\right)\)
Explanation:
(C) : Given, \(\mathrm{x}=\sin ^{-1}(\cos \theta)\)
\(\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{1}{\sqrt{1-\cos ^{2} \theta}} \frac{\mathrm{d}}{\mathrm{d} \theta}(\cos \theta) \Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{-\sin \theta}{\sin \theta} \Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=-1\)
And, \(y=\tan ^{-1} \theta\)
\(\frac{\mathrm{dy}}{\mathrm{d} \theta}=\frac{1}{1+\theta^{2}}\)
Therefore,
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}, \quad \frac{d y}{d x}=\frac{\frac{1}{1+\theta^{2}}}{-1} \Rightarrow \frac{d y}{d x}=-\left(\frac{1}{1+\theta^{2}}\right)\)