Differentiate the following w.r.t. x :
`xsec^(-1)x`.

Answer

To differentiate the function \( y = x \sec^{-1}(x) \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if you have a function \( y = u \cdot v \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

In our case, let:
- \( u = x \)
- \( v = \sec^{-1}(x) \)

Now, we will find the derivatives of \( u \) and \( v \):

1. **Differentiate \( u \)**:
\[ \frac{du}{dx} = 1 \]

2. **Differentiate \( v \)**:
The derivative of \( \sec^{-1}(x) \) is given by:
\[ \frac{dv}{dx} = \frac{1}{|x| \sqrt{x^2 - 1}} \]

Now, we can apply the product rule:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Substituting the values we found:

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

This simplifies to:

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

Thus, the final result for the derivative of \( y = x \sec^{-1}(x) \) is:

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