Differentiate the following w.r.t. x :
`e^(ax)/(sin(bx+c))`

Answer

To differentiate the function \( y = \frac{e^{ax}}{\sin(bx + c)} \) with respect to \( x \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = e^{ax} \) and \( v = \sin(bx + c) \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = e^{ax} \)
- \( v = \sin(bx + c) \)

### Step 2: Differentiate \( u \) and \( v \)
Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

1. Differentiate \( u \):
\[ \frac{du}{dx} = \frac{d}{dx}(e^{ax}) = ae^{ax} \]
(using the chain rule)

2. Differentiate \( v \):
\[ \frac{dv}{dx} = \frac{d}{dx}(\sin(bx + c)) = b \cos(bx + c) \]
(using the chain rule)

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

\[ \frac{dy}{dx} = \frac{\sin(bx + c) \cdot ae^{ax} - e^{ax} \cdot b \cos(bx + c)}{\sin^2(bx + c)} \]

### Step 4: Simplify the Expression
We can factor out \( e^{ax} \) from the numerator:

\[ \frac{dy}{dx} = \frac{e^{ax} \left( a \sin(bx + c) - b \cos(bx + c) \right)}{\sin^2(bx + c)} \]

### Final Result
Thus, the derivative of \( y = \frac{e^{ax}}{\sin(bx + c)} \) with respect to \( x \) is:

\[ \frac{dy}{dx} = \frac{e^{ax} \left( a \sin(bx + c) - b \cos(bx + c) \right)}{\sin^2(bx + c)} \]