Question

The derivative of $\sin^{-1}\!\left(\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ with respect to $\cos^{-1}x$ is

• A. $\dfrac{1}{2}$
• B. $-\dfrac{1}{2}$
• C. $-1$
• D. $1$

Hint:

For derivatives with respect to another function, always use the ratio $\dfrac{dy/dx}{du/dx}$.

Answer 1

The Correct Option is A

Step 1: Simplifying the given expression.
Let \[ y=\sin^{-1}\!\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) \] Using the identity \[ \sin^{-1}\!\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) = \frac{\pi}{4}+\frac{1}{2}\sin^{-1}x \]
Step 2: Differentiating with respect to $x$.
\[ \frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{\sqrt{1-x^2}} \] Also, \[ \frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}} \]
Step 3: Finding the required derivative.
\[ \frac{dy}{d(\cos^{-1}x)} =\frac{\frac{dy}{dx}}{\frac{d}{dx}(\cos^{-1}x)} =\frac{\frac{1}{2\sqrt{1-x^2}}}{-\frac{1}{\sqrt{1-x^2}}} =\frac{1}{2} \]
Step 4: Conclusion.
The required derivative is $\dfrac{1}{2}$.