The derivative of `"tan"^(-1) (2x)` w.r.t `x `is :

Options

A. `1/(1+4x^2)`
B. `2/(1+4x^2)`
C. `2/(sqrt(1-4x^2))`
D. `(-2)/(sqrt(1-4x^2))`

Answer

To find the derivative of \( \tan^{-1}(2x) \) with respect to \( x \), we will use the chain rule. Here’s the step-by-step solution:

### Step 1: Identify the function
We have the function \( y = \tan^{-1}(2x) \).

### Step 2: Apply the chain rule
The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]
In our case, let \( u = 2x \). Thus, \( y = \tan^{-1}(u) \).

### Step 3: Differentiate the outer function
The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = \frac{1}{1 + u^2} \]

### Step 4: Differentiate the inner function
Next, we differentiate \( u = 2x \) with respect to \( x \):
\[ \frac{du}{dx} = 2 \]

### Step 5: Combine the derivatives
Now, applying the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{1 + u^2} \cdot 2 \]
Substituting \( u = 2x \):
\[ \frac{dy}{dx} = \frac{2}{1 + (2x)^2} \]

### Step 6: Simplify the expression
Now, simplify \( (2x)^2 \):
\[ (2x)^2 = 4x^2 \]
Thus, the derivative becomes:
\[ \frac{dy}{dx} = \frac{2}{1 + 4x^2} \]

### Final Answer
The derivative of \( \tan^{-1}(2x) \) with respect to \( x \) is:
\[ \frac{2}{1 + 4x^2} \]

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