Derivative of `sec^(2)(tan^(-1)x)` w.r.t. x is

Options

A. `2x`
B. `x^(2)+1`
C. `x+1`
D. `x^(2)`

Answer

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

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

### Step 2: Use the chain rule
To differentiate \( y \) with respect to \( x \), we apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{d(\tan^{-1}(x))} \cdot \frac{d(\tan^{-1}(x))}{dx} \]

### Step 3: Differentiate \( \sec^2(u) \)
Let \( u = \tan^{-1}(x) \). The derivative of \( \sec^2(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = 2\sec^2(u) \cdot \frac{du}{dx} \]

### Step 4: Differentiate \( \tan^{-1}(x) \)
The derivative of \( \tan^{-1}(x) \) with respect to \( x \) is:
\[ \frac{du}{dx} = \frac{1}{1 + x^2} \]

### Step 5: Substitute back into the chain rule
Now we can substitute \( \frac{du}{dx} \) back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = 2\sec^2(\tan^{-1}(x)) \cdot \frac{1}{1 + x^2} \]

### Step 6: Simplify \( \sec^2(\tan^{-1}(x)) \)
Recall that \( \sec^2(\theta) = 1 + \tan^2(\theta) \). Since \( \theta = \tan^{-1}(x) \), we have:
\[ \tan(\tan^{-1}(x)) = x \implies \sec^2(\tan^{-1}(x)) = 1 + x^2 \]

### Step 7: Substitute \( \sec^2(\tan^{-1}(x)) \) into the derivative
Now substituting this back into our derivative:
\[ \frac{dy}{dx} = 2(1 + x^2) \cdot \frac{1}{1 + x^2} \]

### Step 8: Simplify the expression
The \( (1 + x^2) \) terms cancel out:
\[ \frac{dy}{dx} = 2 \]

### Final Answer
Thus, the derivative of \( \sec^2(\tan^{-1}(x)) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = 2 \]

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