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 solve the problem, we need to find the derivative \(\frac{dy}{dx}\) at \(x = 0\) for the equation:

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

### Step 1: Evaluate \(y\) at \(x = 0\)

First, substitute \(x = 0\) into the equation to find \(y\):

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

Calculating each term:

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

Thus, we have:

\[ y = 1 + 0 = 1 \]

### Step 2: Differentiate both sides with respect to \(x\)

Now, differentiate the original equation implicitly with respect to \(x\):

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

### Step 3: Differentiate the first term

Using the chain rule and product rule, we differentiate \((1 + x)^y\):

Let \(u = (1 + x)^y\).

Using logarithmic differentiation:

\[ \ln(u) = y \ln(1 + x) \]

Differentiating both sides:

\[ \frac{1}{u} \frac{du}{dx} = \frac{dy}{dx} \ln(1 + x) + y \frac{1}{1 + x} \]

Thus,

\[ \frac{du}{dx} = u \left( \frac{dy}{dx} \ln(1 + x) + \frac{y}{1 + x} \right) \]

Substituting back \(u = (1 + x)^y\):

\[ \frac{du}{dx} = (1 + x)^y \left( \frac{dy}{dx} \ln(1 + x) + \frac{y}{1 + x} \right) \]

### Step 4: Differentiate the second term

For the second term \(\sin^{-1}(\sin^2 x)\):

Using the chain rule:

\[ \frac{d}{dx}(\sin^{-1}(\sin^2 x)) = \frac{1}{\sqrt{1 - \sin^4 x}} \cdot 2\sin x \cos x = \frac{\sin(2x)}{\sqrt{1 - \sin^4 x}} \]

### Step 5: Combine the derivatives

Now, we can combine the derivatives:

\[ \frac{dy}{dx} = (1 + x)^y \left( \frac{dy}{dx} \ln(1 + x) + \frac{y}{1 + x} \right) + \frac{\sin(2x)}{\sqrt{1 - \sin^4 x}} \]

### Step 6: Solve for \(\frac{dy}{dx}\)

Rearranging gives:

\[ \frac{dy}{dx} - (1 + x)^y \frac{dy}{dx} \ln(1 + x) = (1 + x)^y \frac{y}{1 + x} + \frac{\sin(2x)}{\sqrt{1 - \sin^4 x}} \]

Factoring out \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} \left( 1 - (1 + x)^y \ln(1 + x) \right) = (1 + x)^{y - 1} y + \frac{\sin(2x)}{\sqrt{1 - \sin^4 x}} \]

Thus,

\[ \frac{dy}{dx} = \frac{(1 + x)^{y - 1} y + \frac{\sin(2x)}{\sqrt{1 - \sin^4 x}}}{1 - (1 + x)^y \ln(1 + x)} \]

### Step 7: Substitute \(x = 0\)

Now substitute \(x = 0\):

\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{(1)^{y - 1} \cdot 1 + 0}{1 - (1)^y \cdot 0} = \frac{1}{1} = 1 \]

### Final Answer

\[ \frac{dy}{dx} \bigg|_{x=0} = 1 \]