To evaluate $f(4)$ for the function $f(x) = x\sqrt{2x + 1}$ by differentiating, we need to follow these steps:

1. Differentiate the function $f(x)$ with respect to $x$ using the product rule.
2. Evaluate the derivative at $x = 4$.

Step 1: Differentiating $f(x) = x \sqrt{2x + 1}$

The function is a product of two parts:

• $x$, and
• $\sqrt{2x + 1}$.

Using the product rule: $(fg)' = f'g + fg'$

Here, let $f(x) = x$ and $g(x) = \sqrt{2x + 1}$.

• The derivative of $x$ is: $\frac{d}{dx} [x] = 1$
• The derivative of $\sqrt{2x + 1}$ is found using the chain rule: [ \frac{d}{dx} \left[ \sqrt{2x + 1} \right] = \frac{1}{2\sqrt{2x + 1}} \cdot \frac{d}{dx} [2x + 1] = \frac{1}{2\sqrt{2x + 1}} \cdot 2 = \frac{1}{\sqrt{2x + 1}} ]

Thus, the derivative of $f(x) = x \sqrt{2x + 1}$ is: $f'(x) = 1 \cdot \sqrt{2x + 1} + x \cdot \frac{1}{\sqrt{2x + 1}}$ Simplifying: $f'(x) = \frac{2x + 1 + x}{\sqrt{2x + 1}} = \frac{3x + 1}{\sqrt{2x + 1}}$

Step 2: Evaluating at $x = 4$

Now substitute $x = 4$ into the derivative: $f'(4) = \frac{3(4) + 1}{\sqrt{2(4) + 1}} = \frac{12 + 1}{\sqrt{8 + 1}} = \frac{13}{\sqrt{9}} = \frac{13}{3}$

Thus, the value of the derivative at $x = 4$ is: $f'(4) = \frac{13}{3}$

Conclusion:

By differentiating, we find that $f'(4) = \frac{13}{3}$.

Would you like more details or explanations on any step?

Related Questions:

1. What is the general rule for differentiating square root functions?
2. How is the product rule applied in more complex cases?
3. Can the chain rule be used directly for a composite function like $\sqrt{2x+1}$?
4. What is the meaning of evaluating the derivative at a point?
5. How do we interpret the result of the derivative in practical applications?

Tip: When applying the chain rule, always differentiate the inner function first before multiplying by the derivative of the outer function.