Differentiate the following w.r.t.x. `(e^(x)+logx)/(sin3x)`

Answer

To differentiate the function \(\frac{e^{x} + \log x}{\sin 3x}\) with respect to \(x\), we will use the quotient rule. The quotient rule states that if you have a function in the form \(\frac{f(x)}{g(x)}\), then its derivative is given by:

\[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2} \]

### Step 1: Identify \(f(x)\) and \(g(x)\)
Let:
- \(f(x) = e^{x} + \log x\)
- \(g(x) = \sin 3x\)

### Step 2: Differentiate \(f(x)\) and \(g(x)\)
Now we need to find \(f'(x)\) and \(g'(x)\).

1. **Differentiate \(f(x)\)**:
- The derivative of \(e^{x}\) is \(e^{x}\).
- The derivative of \(\log x\) is \(\frac{1}{x}\).
- Therefore,
\[ f'(x) = e^{x} + \frac{1}{x} \]

2. **Differentiate \(g(x)\)**:
- The derivative of \(\sin 3x\) using the chain rule is \(3 \cos 3x\).
- Therefore,
\[ g'(x) = 3 \cos 3x \]

### Step 3: Apply the Quotient Rule
Now we can apply the quotient rule:

\[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{(e^{x} + \frac{1}{x}) \sin 3x - (3 \cos 3x)(e^{x} + \log x)}{(\sin 3x)^2} \]

### Step 4: Simplify the Expression
Now, we can simplify the expression:

\[ \frac{d}{dx}\left(\frac{e^{x} + \log x}{\sin 3x}\right) = \frac{(e^{x} + \frac{1}{x}) \sin 3x - 3 \cos 3x (e^{x} + \log x)}{\sin^2 3x} \]

### Final Answer
Thus, the derivative of the given function is:

\[ \frac{(e^{x} + \frac{1}{x}) \sin 3x - 3 \cos 3x (e^{x} + \log x)}{\sin^2 3x} \]

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