Differentiate w.r.t x
`tan^-1((2x)/(1+15x^2))`

Answer

To differentiate the function \( y = \tan^{-1}\left(\frac{2x}{1 + 15x^2}\right) \) with respect to \( x \), we can use the formula for the derivative of the inverse tangent function, which is:

\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

where \( u = \frac{2x}{1 + 15x^2} \).

### Step 1: Differentiate the inner function \( u \)

First, we need to differentiate \( u \):

\[ u = \frac{2x}{1 + 15x^2} \]

Using the quotient rule, which states that if \( u = \frac{f(x)}{g(x)} \), then:

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

Here, \( f(x) = 2x \) and \( g(x) = 1 + 15x^2 \).

Calculating \( f'(x) \) and \( g'(x) \):

\[ f'(x) = 2 \]
\[ g'(x) = 30x \]

Now applying the quotient rule:

\[ \frac{du}{dx} = \frac{(2)(1 + 15x^2) - (2x)(30x)}{(1 + 15x^2)^2} \]

Simplifying the numerator:

\[ = \frac{2 + 30x^2 - 60x^2}{(1 + 15x^2)^2} \]
\[ = \frac{2 - 30x^2}{(1 + 15x^2)^2} \]

### Step 2: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula

Now we substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula:

\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{2x}{1 + 15x^2}\right)^2} \cdot \frac{du}{dx} \]

Calculating \( 1 + u^2 \):

\[ u^2 = \left(\frac{2x}{1 + 15x^2}\right)^2 = \frac{4x^2}{(1 + 15x^2)^2} \]

Thus,

\[ 1 + u^2 = 1 + \frac{4x^2}{(1 + 15x^2)^2} = \frac{(1 + 15x^2)^2 + 4x^2}{(1 + 15x^2)^2} \]

Expanding \( (1 + 15x^2)^2 \):

\[ = 1 + 30x^2 + 225x^4 + 4x^2 = 1 + 34x^2 + 225x^4 \]

So,

\[ 1 + u^2 = \frac{1 + 34x^2 + 225x^4}{(1 + 15x^2)^2} \]

Now substituting back into the derivative:

\[ \frac{dy}{dx} = \frac{(1 + 15x^2)^2}{1 + 34x^2 + 225x^4} \cdot \frac{2 - 30x^2}{(1 + 15x^2)^2} \]

### Step 3: Simplify the expression

The \( (1 + 15x^2)^2 \) cancels out:

\[ \frac{dy}{dx} = \frac{2 - 30x^2}{1 + 34x^2 + 225x^4} \]

Thus, the final answer is:

\[ \frac{dy}{dx} = \frac{2 - 30x^2}{1 + 34x^2 + 225x^4} \]