Derivative of y= sec ` (tan ^(-1) x )` at x=1 is _________

Options

A. ` (1)/(2 `
B. 1
C. ` 1//sqrt2`
D. ` sqrt2 `

Answer

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

### Step 1: Differentiate the outer function
We start by applying the chain rule. The derivative of \( \sec(u) \) with respect to \( u \) is \( \sec(u) \tan(u) \). Here, \( u = \tan^{-1}(x) \).

\[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \tan(\tan^{-1}(x)) \cdot \frac{d}{dx}(\tan^{-1}(x)) \]

### Step 2: Simplify \( \tan(\tan^{-1}(x)) \)
Since \( \tan(\tan^{-1}(x)) = x \), we can substitute this into our derivative expression.

\[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \cdot x \cdot \frac{d}{dx}(\tan^{-1}(x)) \]

### Step 3: Differentiate \( \tan^{-1}(x) \)
The derivative of \( \tan^{-1}(x) \) is given by:

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

### Step 4: Substitute back into the derivative
Now we substitute this back into our expression for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \cdot x \cdot \frac{1}{1 + x^2} \]

### Step 5: Evaluate at \( x = 1 \)
Now we need to evaluate \( \frac{dy}{dx} \) at \( x = 1 \):

\[ \frac{dy}{dx} \bigg|_{x=1} = \sec(\tan^{-1}(1)) \cdot 1 \cdot \frac{1}{1 + 1^2} \]

### Step 6: Calculate \( \tan^{-1}(1) \)
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \).

### Step 7: Calculate \( \sec(\frac{\pi}{4}) \)
The value of \( \sec(\frac{\pi}{4}) \) is:

\[ \sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]

### Step 8: Substitute values
Now we substitute this back into our derivative expression:

\[ \frac{dy}{dx} \bigg|_{x=1} = \sqrt{2} \cdot 1 \cdot \frac{1}{1 + 1} = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2} \]

### Final Answer
Thus, the derivative of \( y = \sec(\tan^{-1}(x)) \) at \( x = 1 \) is:

\[ \frac{\sqrt{2}}{2} \]

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