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

Options

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

Answer

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 \(y\) using the chain rule
We start with the function:
\[ y = \sec(\tan^{-1}(x)) \]
Using the chain rule, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}[\sec(u)] \cdot \frac{du}{dx} \]
where \(u = \tan^{-1}(x)\).

### Step 2: Differentiate \(\sec(u)\)
The derivative of \(\sec(u)\) is:
\[ \frac{d}{du}[\sec(u)] = \sec(u) \tan(u) \]
Thus,
\[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \tan(\tan^{-1}(x)) \cdot \frac{du}{dx} \]

### Step 3: Differentiate \(u = \tan^{-1}(x)\)
The derivative of \(\tan^{-1}(x)\) is:
\[ \frac{du}{dx} = \frac{1}{1 + x^2} \]
Substituting this back into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \tan(\tan^{-1}(x)) \cdot \frac{1}{1 + x^2} \]

### Step 4: Simplify \(\tan(\tan^{-1}(x))\)
Since \(\tan(\tan^{-1}(x)) = x\), we have:
\[ \frac{dy}{dx} = \sec(\tan^{-1}(x)) \cdot x \cdot \frac{1}{1 + x^2} \]

### Step 5: Find \(\sec(\tan^{-1}(x))\)
To find \(\sec(\tan^{-1}(x))\), we can use the identity:
\[ \sec(\theta) = \frac{1}{\cos(\theta)} \]
where \(\theta = \tan^{-1}(x)\). In a right triangle where the opposite side is \(x\) and the adjacent side is \(1\), the hypotenuse \(h\) is:
\[ h = \sqrt{x^2 + 1} \]
Thus,
\[ \sec(\tan^{-1}(x)) = \frac{h}{1} = \sqrt{x^2 + 1} \]

### Step 6: Substitute back into \(\frac{dy}{dx}\)
Now we can substitute \(\sec(\tan^{-1}(x))\) 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 7: 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:
\[ \frac{\sqrt{2}}{2} \]

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