If `y= sec (tan^(-1)x)," then "(dy)/(dx)" at " x = 1` is equal to

Options

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

Explanation

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

### Step 1: Differentiate the function
We start with the function:
\[ y = \sec(\tan^{-1}(x)) \]
To differentiate this, we will use the chain rule. The derivative of \(\sec(u)\) is \(\sec(u) \tan(u) \frac{du}{dx}\), where \(u = \tan^{-1}(x)\).

### Step 2: Find \(\frac{du}{dx}\)
First, we find \(\frac{du}{dx}\):
\[ u = \tan^{-1}(x) \implies \frac{du}{dx} = \frac{1}{1 + x^2} \]

### Step 3: Apply the chain rule
Now, applying the chain rule:
\[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \tan(\tan^{-1}(x)) \cdot \frac{du}{dx} \]
Since \(\tan(\tan^{-1}(x)) = x\), we can substitute this into our equation:
\[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \cdot x \cdot \frac{1}{1 + x^2} \]

### Step 4: Simplify \(\sec(\tan^{-1}(x))\)
Next, we need to find \(\sec(\tan^{-1}(x))\). From the definition of \(\tan^{-1}(x)\), we can visualize a right triangle where:
- The opposite side is \(x\)
- The adjacent side is \(1\)
- The hypotenuse is \(\sqrt{x^2 + 1}\)

Thus, we have:
\[ \sec(\tan^{-1}(x)) = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{\sqrt{x^2 + 1}}{1} = \sqrt{x^2 + 1} \]

### Step 5: Substitute back into the derivative
Now substituting back into our derivative:
\[ \frac{dy}{dx} = \sqrt{x^2 + 1} \cdot x \cdot \frac{1}{1 + x^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{x \sqrt{x^2 + 1}}{1 + x^2} \]

### Step 6: Evaluate at \(x = 1\)
Now we evaluate \(\frac{dy}{dx}\) at \(x = 1\):
\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{1 \cdot \sqrt{1^2 + 1}}{1 + 1^2} = \frac{1 \cdot \sqrt{2}}{2} = \frac{\sqrt{2}}{2} \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(x = 1\) is:
\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{\sqrt{2}}{2} \]

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