If ` y = sin[cos^(-1){sin(cos^(-1) x)}], "then" (dy)/(dx)" at x" = (1)/(2)`

Options

A. 0
B. -1
C. `(2)/(sqrt(3))`
D. 1

Explanation

To solve the problem step by step, we start with the given function:

**Step 1: Rewrite the function.**
Given:
\[ y = \sin\left(\cos^{-1}\left(\sin\left(\cos^{-1}(x)\right)\right)\right) \]

We can simplify this expression using the identity:
\[ \cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x) \]

So, we can rewrite:
\[ \cos^{-1}(x) = \frac{\pi}{2} - \theta \quad \text{where } \theta = \sin^{-1}(x) \]

Thus:
\[ \sin\left(\cos^{-1}(x)\right) = \sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta) = \sqrt{1 - x^2} \]

Now substituting back:
\[ y = \sin\left(\cos^{-1}(\sin(\theta))\right) \]

**Step 2: Simplify further.**
Since \(\sin(\theta) = x\), we have:
\[ y = \sin\left(\cos^{-1}(x)\right) \]

Using the identity again:
\[ \sin\left(\cos^{-1}(x)\right) = \sqrt{1 - x^2} \]

So now:
\[ y = \sqrt{1 - x^2} \]

**Step 3: Differentiate y with respect to x.**
Now we differentiate \(y\):
\[ \frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 - x^2}) \]

Using the chain rule:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]

**Step 4: Evaluate at \(x = \frac{1}{2}\).**
Now we need to find \(\frac{dy}{dx}\) at \(x = \frac{1}{2}\):
\[ \frac{dy}{dx}\bigg|_{x = \frac{1}{2}} = \frac{-\frac{1}{2}}{\sqrt{1 - \left(\frac{1}{2}\right)^2}} \]
\[ = \frac{-\frac{1}{2}}{\sqrt{1 - \frac{1}{4}}} = \frac{-\frac{1}{2}}{\sqrt{\frac{3}{4}}} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{-1}{\sqrt{3}} \]

Thus, the final answer is:
\[ \frac{dy}{dx} \text{ at } x = \frac{1}{2} = \frac{-1}{\sqrt{3}} \]

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