Question
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\},0<\text{x}<1$
$\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\},0<\text{x}<1$
✓
Answer
Let $\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1+\cos2\theta}{2}}\Big\}$
$=\sin^{-1}\Big\{\sqrt{\frac{2\sin^2\theta}{2}}\Big\}$
$\text{y}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
$\Rightarrow\ 0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\theta$
$\Big[\text{Since, } \sin^{-1}(\sin\theta)=\theta\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}\ \big[\text{Since x}=\cos2\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{1-\text{x}^2}}$
Put $\text{x}=\cos2\theta$
$\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1+\cos2\theta}{2}}\Big\}$
$=\sin^{-1}\Big\{\sqrt{\frac{2\sin^2\theta}{2}}\Big\}$
$\text{y}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
$\Rightarrow\ 0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\theta$
$\Big[\text{Since, } \sin^{-1}(\sin\theta)=\theta\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}\ \big[\text{Since x}=\cos2\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{1-\text{x}^2}}$
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