Find $$ \frac{d y}{d x} $$ of the following functions; (i) $$ x \ln (\operatorname{Sin} x)=y $$ (ii) $$ f(x)=e^{\operatorname{Sin} x^{2}} $$

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**(i) Given:**  
$$
y = x \ln (\sin x)
$$

**Step 1: Differentiate using the product rule.**  
The product rule states that if $$ y = u(x)v(x) $$, then  
$$
\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x).
$$  
Let $$ u(x) = x $$ and $$ v(x) = \ln (\sin x) $$.

**Step 2: Compute the derivatives.**  
- For $$ u(x) = x $$, we have  
  $$
  u'(x) = 1.
  $$
- For $$ v(x) = \ln (\sin x) $$, use the chain rule. The derivative of $$ \ln (w) $$ with respect to $$ w $$ is $$ \frac{1}{w} $$ and then multiplied by the derivative of $$ w $$ (where $$ w = \sin x $$). Thus,  
  $$
  v'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x.
  $$

**Step 3: Apply the product rule.**  
$$
\frac{dy}{dx} = 1 \cdot \ln (\sin x) + x \cdot \cot x = \ln (\sin x) + x \cot x.
$$

---

**(ii) Given:**  
$$
f(x) = e^{\sin (x^2)}
$$

**Step 1: Differentiate using the chain rule.**  
Let $$ u(x) = \sin (x^2) $$ so that $$ f(x) = e^{u(x)} $$. The derivative of $$ e^{u(x)} $$ is  
$$
f'(x) = e^{u(x)} \cdot u'(x).
$$

**Step 2: Differentiate $$ u(x) = \sin (x^2) $$.**  
Apply the chain rule again. Let $$ v(x) = x^2 $$ so that $$ u(x) = \sin (v(x)) $$.  
- The derivative of $$ \sin (v) $$ with respect to $$ v $$ is $$ \cos (v) $$.  
- The derivative of $$ v(x) = x^2 $$ is $$ 2x $$.

Thus,  
$$
u'(x) = \cos (x^2) \cdot 2x = 2x \cos (x^2).
$$

**Step 3: Combine the results.**  
$$
f'(x) = e^{\sin (x^2)} \cdot 2x \cos (x^2) = 2x \cos (x^2) e^{\sin (x^2)}.
$$
**(i) $$ \frac{dy}{dx} = \ln (\sin x) + x \cot x $$** **(ii) $$ f'(x) = 2x \cos (x^2) e^{\sin (x^2)} $$**

Find \( \frac{d y}{d x} \) of the following functions; (i) \( x \ln (\operatorname{Sin} x)=y \) (ii) \( f(x)=e^{\operatorname{Sin} x^{2}} \)

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