Differentiate the following functions w.r.ts
(i) ` e^(x^(2)tan^(-1)x)/(sqrt(1-x^(2))`
(ii) ` x^(sinx) " " (x gt 0)`

Answer

To differentiate the given functions step by step, we will follow the standard rules of differentiation including the product rule, quotient rule, and chain rule.

### (i) Differentiate \( y = \frac{e^{x^2 \tan^{-1} x}}{\sqrt{1 - x^2}} \)

1. **Identify the function**: We have a quotient of two functions: the numerator \( u = e^{x^2 \tan^{-1} x} \) and the denominator \( v = \sqrt{1 - x^2} \).

2. **Apply the quotient rule**: The derivative of \( y = \frac{u}{v} \) is given by:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

3. **Differentiate the numerator \( u \)**:
- Using the chain rule:
\[ \frac{du}{dx} = e^{x^2 \tan^{-1} x} \cdot \frac{d}{dx}(x^2 \tan^{-1} x) \]
- Differentiate \( x^2 \tan^{-1} x \) using the product rule:
\[ \frac{d}{dx}(x^2 \tan^{-1} x) = 2x \tan^{-1} x + x^2 \cdot \frac{1}{1 + x^2} \]
- Thus,
\[ \frac{du}{dx} = e^{x^2 \tan^{-1} x} \left( 2x \tan^{-1} x + \frac{x^2}{1 + x^2} \right) \]

4. **Differentiate the denominator \( v \)**:
- Using the chain rule:
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]

5. **Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule**:
\[ \frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot e^{x^2 \tan^{-1} x} \left( 2x \tan^{-1} x + \frac{x^2}{1 + x^2} \right) - e^{x^2 \tan^{-1} x} \cdot \left( \frac{-x}{\sqrt{1 - x^2}} \right)}{(1 - x^2)} \]

6. **Simplify the expression**:
- Combine the terms in the numerator and simplify as needed.

### Final Result for (i):
\[ \frac{dy}{dx} = \frac{e^{x^2 \tan^{-1} x}}{(1 - x^2)} \left( \sqrt{1 - x^2} \left( 2x \tan^{-1} x + \frac{x^2}{1 + x^2} \right) + x \right) \]

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### (ii) Differentiate \( y = x^{\sin x} \)

1. **Identify the function**: We have \( y = x^{\sin x} \).

2. **Take the natural logarithm**:
\[ \ln y = \sin x \cdot \ln x \]

3. **Differentiate both sides**:
- Using implicit differentiation:
\[ \frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \]

4. **Solve for \( \frac{dy}{dx} \)**:
\[ \frac{dy}{dx} = y \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]

5. **Substitute back for \( y \)**:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]

### Final Result for (ii):
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]

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