What is the slope of the tangent to the curve `y=sin^(-1)(sin^(2)x)` at x=0?

Options

A. 0
B. 1
C. 2
D. None of these

Answer

To find the slope of the tangent to the curve \( y = \sin^{-1}(\sin^2 x) \) at \( x = 0 \), we will follow these steps:

### Step 1: Define the function
Let \( y = \sin^{-1}(\sin^2 x) \).

### Step 2: Differentiate the function
To find the slope of the tangent, we need to compute the derivative \( \frac{dy}{dx} \). We will use the chain rule for differentiation.

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

### Step 3: Differentiate \( \sin^2 x \)
Now, we need to differentiate \( \sin^2 x \):
\[ \frac{d}{dx}(\sin^2 x) = 2 \sin x \cdot \cos x = \sin(2x) \]

### Step 4: Substitute back into the derivative
Substituting back into the derivative:
\[ \frac{dy}{dx} = \frac{\sin(2x)}{\sqrt{1 - \sin^4 x}} \]

### Step 5: Evaluate the derivative at \( x = 0 \)
Now, we will evaluate \( \frac{dy}{dx} \) at \( x = 0 \):
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{\sin(2 \cdot 0)}{\sqrt{1 - \sin^4(0)}} \]
Since \( \sin(0) = 0 \):
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{0}{\sqrt{1 - 0}} = \frac{0}{1} = 0 \]

### Conclusion
Thus, the slope of the tangent to the curve at \( x = 0 \) is \( 0 \).

### Final Answer
The slope of the tangent to the curve \( y = \sin^{-1}(\sin^2 x) \) at \( x = 0 \) is \( 0 \).

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