The problem states:

Let $g(x) = x^2 \sqrt{f(x)}$, where $f$ is a differentiable function such that $f(1) = 4$ and $f'(1) = 1$. Find $g'(1)$.

Solution:

We will differentiate $g(x)$ using the product rule and chain rule.

Step 1: Rewrite $g(x)$

The function is: $g(x) = x^2 \sqrt{f(x)} = x^2 \cdot f(x)^{1/2}.$

Step 2: Differentiate $g(x)$

Using the product rule, the derivative is: [ g'(x) = \frac{d}{dx}[x^2] \cdot f(x)^{1/2} + x^2 \cdot \frac{d}{dx}[f(x)^{1/2}]. ]

• The derivative of $x^2$ is $2x$, so: $\frac{d}{dx}[x^2] \cdot f(x)^{1/2} = 2x \cdot f(x)^{1/2}.$

• The derivative of $f(x)^{1/2}$ (using the chain rule) is: $\frac{d}{dx}[f(x)^{1/2}] = \frac{1}{2} f(x)^{-1/2} \cdot f'(x).$ Thus: $x^2 \cdot \frac{d}{dx}[f(x)^{1/2}] = x^2 \cdot \frac{1}{2} f(x)^{-1/2} \cdot f'(x).$

Combining these, we get: $g'(x) = 2x f(x)^{1/2} + x^2 \cdot \frac{1}{2} f(x)^{-1/2} \cdot f'(x).$

Step 3: Evaluate $g'(1)$

Substitute $x = 1$, $f(1) = 4$, and $f'(1) = 1$ into the derivative: $g'(1) = 2(1) \cdot \sqrt{f(1)} + (1)^2 \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{f(1)}} \cdot f'(1).$

• $\sqrt{f(1)} = \sqrt{4} = 2$,
• $\frac{1}{\sqrt{f(1)}} = \frac{1}{2}$,
• $f'(1) = 1$.

Now compute each term: $2(1) \cdot \sqrt{f(1)} = 2 \cdot 2 = 4,$ $(1)^2 \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{f(1)}} \cdot f'(1) = 1 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 1 = \frac{1}{4}.$

Add these: $g'(1) = 4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4}.$

Final Answer:

$\boxed{\frac{17}{4}}$

Would you like more details on this solution or further clarifications?

5 Related Questions:

1. What are the general steps for applying the product rule to composite functions?
2. How does the chain rule work when differentiating a square root function?
3. Can you explain why $f(x)^{1/2}$ differentiates to $\frac{1}{2} f(x)^{-1/2} \cdot f'(x)$?
4. How can the values of $f'(1)$ and $f(1)$ be used to simplify derivatives involving $f(x)$?
5. What happens if we had a higher power for $x$ in $g(x) = x^n \sqrt{f(x)}$?

Tip:

When dealing with derivatives of composite functions, carefully rewrite the function to apply both the product and chain rules systematically.