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

Options

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

Answer

To find the derivative of \( y = \sec(\tan^{-1} x) \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \sec(\tan^{-1} x) \]

### Step 2: Use the identity for \( \tan^{-1} x \)
Recall that:
\[ \tan^{-1} x = \theta \implies x = \tan \theta \]
This means:
\[ \sec(\tan^{-1} x) = \sec(\theta) \]

### Step 3: Relate secant to the right triangle
From the definition of secant:
\[ \sec(\theta) = \frac{1}{\cos(\theta)} \]
And using the right triangle relationships:
\[ \sec(\theta) = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{1 + x^2}}{1} \]
Thus:
\[ y = \sqrt{1 + x^2} \]

### Step 4: Differentiate \( y \)
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 + x^2}) \]

Using the chain rule:
\[ \frac{dy}{dx} = \frac{1}{2}(1 + x^2)^{-1/2} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}} \]

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

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