If `y=(1+x)^y+sin^-1(sin^2x)` , then `(dy)/(dx)` at x = 0 is

Options

A. 0
B. In 2
C. 1
D. `1/2`

Answer

To find \(\frac{dy}{dx}\) at \(x = 0\) for the equation \(y = (1+x)^y + \sin^{-1}(\sin^2 x)\), we will follow these steps:

### Step 1: Evaluate \(y\) at \(x = 0\)
First, we substitute \(x = 0\) into the equation to find the value of \(y\).

\[ y = (1+0)^y + \sin^{-1}(\sin^2(0)) \]
\[ y = 1^y + \sin^{-1}(0) \]
\[ y = 1 + 0 \]
\[ y = 1 \]

### Step 2: Differentiate both sides with respect to \(x\)
Now we differentiate the equation \(y = (1+x)^y + \sin^{-1}(\sin^2 x)\) with respect to \(x\).

Using implicit differentiation:
\[ \frac{dy}{dx} = \frac{d}{dx}[(1+x)^y] + \frac{d}{dx}[\sin^{-1}(\sin^2 x)] \]

### Step 3: Differentiate \((1+x)^y\)
Using the chain rule and product rule:
\[ \frac{d}{dx}[(1+x)^y] = y(1+x)^{y-1} \cdot \frac{d}{dx}(1+x) + (1+x)^y \cdot \frac{dy}{dx} \cdot \ln(1+x) \]
\[ = y(1+x)^{y-1} + (1+x)^y \ln(1+x) \frac{dy}{dx} \]

### Step 4: Differentiate \(\sin^{-1}(\sin^2 x)\)
Using the chain rule:
\[ \frac{d}{dx}[\sin^{-1}(\sin^2 x)] = \frac{1}{\sqrt{1 - \sin^4 x}} \cdot 2\sin x \cos x \]
\[ = \frac{2\sin x \cos x}{\sqrt{1 - \sin^4 x}} \]

### Step 5: Combine the derivatives
Now we combine the derivatives:
\[ \frac{dy}{dx} = y(1+x)^{y-1} + (1+x)^y \ln(1+x) \frac{dy}{dx} + \frac{2\sin x \cos x}{\sqrt{1 - \sin^4 x}} \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Rearranging gives:
\[ \frac{dy}{dx} - (1+x)^y \ln(1+x) \frac{dy}{dx} = y(1+x)^{y-1} + \frac{2\sin x \cos x}{\sqrt{1 - \sin^4 x}} \]
\[ \frac{dy}{dx}(1 - (1+x)^y \ln(1+x)) = y(1+x)^{y-1} + \frac{2\sin x \cos x}{\sqrt{1 - \sin^4 x}} \]
\[ \frac{dy}{dx} = \frac{y(1+x)^{y-1} + \frac{2\sin x \cos x}{\sqrt{1 - \sin^4 x}}}{1 - (1+x)^y \ln(1+x)} \]

### Step 7: Substitute \(x = 0\) into the derivative
Now we substitute \(x = 0\) into the expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1 \cdot 1^{1-1} + \frac{2 \cdot 0 \cdot 1}{\sqrt{1 - 0}}}{1 - 1 \cdot \ln(1)} \]
\[ = \frac{1 + 0}{1 - 0} \]
\[ = 1 \]

### Final Answer
Thus, \(\frac{dy}{dx}\) at \(x = 0\) is \(1\).