Question

hw11 differentiation rules ii (target c1, c2, c5; §3.3)
score: 5/8 answered: 5/8
question 6
if (f(x)=\frac{sqrt{x}-5}{sqrt{x}+5}), find:
(f(x)=)
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Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \sqrt{x}-5=x^{\frac{1}{2}}-5$, $v=\sqrt{x}+5=x^{\frac{1}{2}}+5$.

Step2: Find $u^\prime$ and $v^\prime$

Differentiate $u$ with respect to $x$: $u^\prime=\frac{d}{dx}(x^{\frac{1}{2}}-5)=\frac{1}{2}x^{-\frac{1}{2}}$. Differentiate $v$ with respect to $x$: $v^\prime=\frac{d}{dx}(x^{\frac{1}{2}} + 5)=\frac{1}{2}x^{-\frac{1}{2}}$.

Step3: Substitute into quotient - rule formula

$f^\prime(x)=\frac{(\frac{1}{2}x^{-\frac{1}{2}})(\sqrt{x}+5)-(\sqrt{x}-5)(\frac{1}{2}x^{-\frac{1}{2}})}{(\sqrt{x}+5)^{2}}$.
Simplify the numerator:
\[

$$\begin{align*} &(\frac{1}{2}x^{-\frac{1}{2}})(\sqrt{x}+5)-(\sqrt{x}-5)(\frac{1}{2}x^{-\frac{1}{2}})\\ =&\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}+\frac{5}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}+\frac{5}{2}x^{-\frac{1}{2}}\\ =&\frac{5}{2}x^{-\frac{1}{2}}+\frac{5}{2}x^{-\frac{1}{2}}\\ =&5x^{-\frac{1}{2}} \end{align*}$$

\]
So, $f^\prime(x)=\frac{5x^{-\frac{1}{2}}}{(\sqrt{x}+5)^{2}}=\frac{5}{(\sqrt{x}+5)^{2}\sqrt{x}}$.

Answer:

$\frac{5}{(\sqrt{x}+5)^{2}\sqrt{x}}$