Differentiate the following w.r.t.x :
`(sin(ax+b))/(cos(cx+d))`

Answer

To differentiate the function \( \frac{\sin(ax + b)}{\cos(cx + d)} \) with respect to \( x \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function \( \frac{U}{V} \), then the derivative is given by:

\[ \frac{d}{dx} \left( \frac{U}{V} \right) = \frac{V \frac{dU}{dx} - U \frac{dV}{dx}}{V^2} \]

Here, let:
- \( U = \sin(ax + b) \)
- \( V = \cos(cx + d) \)

### Step 1: Differentiate \( U \) and \( V \)

First, we need to find \( \frac{dU}{dx} \) and \( \frac{dV}{dx} \).

1. **Differentiate \( U = \sin(ax + b) \)**:
\[ \frac{dU}{dx} = a \cos(ax + b) \]

2. **Differentiate \( V = \cos(cx + d) \)**:
\[ \frac{dV}{dx} = -c \sin(cx + d) \]

### Step 2: Apply the Quotient Rule

Now, we can substitute \( U \), \( V \), \( \frac{dU}{dx} \), and \( \frac{dV}{dx} \) into the quotient rule formula:

\[ \frac{d}{dx} \left( \frac{\sin(ax + b)}{\cos(cx + d)} \right) = \frac{\cos(cx + d) \cdot (a \cos(ax + b)) - \sin(ax + b) \cdot (-c \sin(cx + d))}{(\cos(cx + d))^2} \]

### Step 3: Simplify the Expression

Now, simplify the expression:

\[ = \frac{a \cos(ax + b) \cos(cx + d) + c \sin(ax + b) \sin(cx + d)}{(\cos(cx + d))^2} \]

### Final Answer

Thus, the derivative of \( \frac{\sin(ax + b)}{\cos(cx + d)} \) with respect to \( x \) is:

\[ \frac{a \cos(ax + b) \cos(cx + d) + c \sin(ax + b) \sin(cx + d)}{(\cos(cx + d))^2} \]