7. Calculate $$ f^{\prime}(1) $$ for $$ f(x)=\frac{x^{4}+x^{3}+x^{2}+x+1}{\sqrt{x}+1} $$

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UpStudy AI · Accepted Answer

**Step 1. Identify the function components**

Given
$$
f(x)=\frac{N(x)}{D(x)} \quad \text{where} \quad N(x)=x^4+x^3+x^2+x+1, \quad D(x)=\sqrt{x}+1.
$$

**Step 2. Compute the derivatives of $$N(x)$$ and $$D(x)$$**

Differentiate $$N(x)$$ term by term:
$$
N'(x) = 4x^3+3x^2+2x+1.
$$

Differentiate $$D(x)=\sqrt{x}+1$$. Recall that $$\sqrt{x}=x^{1/2}$$, so:
$$
D'(x)=\frac{d}{dx}(x^{1/2})+0=\frac{1}{2\sqrt{x}}.
$$

**Step 3. Evaluate the functions and their derivatives at $$x=1$$**

Substitute $$x=1$$ into $$N(x)$$ and $$N'(x)$$:
$$
N(1)=1^4+1^3+1^2+1+1=5,
$$
$$
N'(1)=4(1)^3+3(1)^2+2(1)+1=4+3+2+1=10.
$$

Similarly, evaluate $$D(x)$$ and $$D'(x)$$ at $$x=1$$:
$$
D(1)=\sqrt{1}+1=1+1=2,
$$
$$
D'(1)=\frac{1}{2\sqrt{1}}=\frac{1}{2}.
$$

**Step 4. Apply the Quotient Rule**

The derivative of $$f(x)$$ is given by:
$$
f'(x) = \frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^2}.
$$

Substitute the values at $$x=1$$:
$$
f'(1)=\frac{10\cdot2-5\cdot\frac{1}{2}}{2^2}.
$$

**Step 5. Simplify the expression**

First, compute the numerator:
$$
10\cdot 2=20 \quad \text{and} \quad 5\cdot\frac{1}{2}=2.5.
$$
Thus,
$$
20-2.5=17.5.
$$
The denominator is:
$$
2^2=4.
$$
So,
$$
f'(1)=\frac{17.5}{4}.
$$

Express $$17.5$$ as a fraction:
$$
17.5=\frac{35}{2},
$$
therefore,
$$
f'(1)=\frac{\frac{35}{2}}{4}=\frac{35}{8}.
$$

**Final Answer**

$$
f'(1)=\frac{35}{8}.
$$
$$ f'(1) = \frac{35}{8} $$

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