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

Options

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

Answer

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

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

### Step 2: Find \( \frac{du}{dx} \)
Now, we need to differentiate \( u = \tan^{-1} x \):
\[ \frac{du}{dx} = \frac{1}{1 + x^2} \]

### Step 3: Substitute back into the derivative
Now we can substitute \( u \) back into the derivative:
\[ \frac{dy}{dx} = \sec(\tan^{-1} x) \tan(\tan^{-1} x) \cdot \frac{1}{1 + x^2} \]

### Step 4: Simplify \( \sec(\tan^{-1} x) \) and \( \tan(\tan^{-1} x) \)
We know that:
- \( \tan(\tan^{-1} x) = x \)
- To find \( \sec(\tan^{-1} x) \), we can use the identity \( \sec(\theta) = \sqrt{1 + \tan^2(\theta)} \). Thus:
\[ \sec(\tan^{-1} x) = \sqrt{1 + x^2} \]

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

### Step 6: Final simplification
This simplifies to:
\[ \frac{dy}{dx} = \frac{x \sqrt{1 + x^2}}{1 + x^2} \]

### Conclusion
Thus, the final result for the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{1 + x^2}} \]

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