We are tasked with finding the derivative of the given function:

$y(x) = \sqrt{x^2 + 1}$

Step-by-step process:

1. Rewrite the square root in exponent form: $y(x) = (x^2 + 1)^{1/2}$

2. Differentiate using the chain rule: The chain rule states that: $\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$ Here, $f(u) = u^{1/2}$ and $g(x) = x^2 + 1$. We need to differentiate both parts.

   • Derivative of the outer function $f(u) = u^{1/2}$: $\frac{d}{du} [u^{1/2}] = \frac{1}{2} u^{-1/2}$

   • Derivative of the inner function $g(x) = x^2 + 1$: $\frac{d}{dx} [x^2 + 1] = 2x$

3. Apply the chain rule: $\frac{dy}{dx} = \frac{1}{2} (x^2 + 1)^{-1/2} \cdot 2x$

4. Simplify the expression: The 2's cancel out: $\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}}$

Thus, the derivative is:

$\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}}$

Let me know if you would like more details or have any questions!

Related questions:

1. How would you differentiate $y = \sqrt{2x + 3}$?
2. Can you explain the chain rule in more detail?
3. What is the derivative of $y = \ln(x^2 + 1)$?
4. How do you apply the product rule in differentiation?
5. What happens when differentiating functions involving trigonometric terms?

Tip:

Always check if rewriting expressions (like square roots) in exponent form can make differentiation easier.