Differentiate the following functions w.r.t. x:
`(tan x)/( 2x +3)`

Answer

To differentiate the function \( y = \frac{\tan x}{2x + 3} \) with respect to \( x \), we will use the quotient rule. The quotient rule states that if you have a function in the form \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

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

Here, we will identify \( u \) and \( v \) as follows:
- \( u = \tan x \)
- \( v = 2x + 3 \)

### Step 1: Differentiate \( u \) and \( v \)
First, we need to find the derivatives of \( u \) and \( v \):
- The derivative of \( u = \tan x \) is \( \frac{du}{dx} = \sec^2 x \).
- The derivative of \( v = 2x + 3 \) is \( \frac{dv}{dx} = 2 \).

### Step 2: Apply the Quotient Rule
Now, we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{(2x + 3)(\sec^2 x) - (\tan x)(2)}{(2x + 3)^2} \]

### Step 3: Simplify the Expression
We will simplify the expression in the numerator:

\[ \frac{dy}{dx} = \frac{(2x + 3)\sec^2 x - 2\tan x}{(2x + 3)^2} \]

### Final Result
Thus, the derivative of the function \( y = \frac{\tan x}{2x + 3} \) with respect to \( x \) is:

\[ \frac{dy}{dx} = \frac{(2x + 3)\sec^2 x - 2\tan x}{(2x + 3)^2} \]

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