Question

homework3: problem 30
(1 point)
let $f(x)=\frac{(5 x^{2}-5)(x^{2}-6)}{x^{2}+6}$. evaluate $f^{prime}(x)$ at $x = 5$.
$f^{prime}(5)=\square$

Explanation:

Step1: Use the quotient rule

The quotient rule states that if \(y=\frac{u}{v}\), then \(y'=\frac{u'v - uv'}{v^{2}}\). Let \(u=(5x^{2}-5)(x^{2}-6)=5x^{4}-30x^{2}-5x^{2}+30 = 5x^{4}-35x^{2}+30\), and \(v=x^{2}+6\).
First, find \(u'\):
\(u'=\frac{d}{dx}(5x^{4}-35x^{2}+30)=20x^{3}-70x\)
And \(v'=\frac{d}{dx}(x^{2}+6)=2x\)

Step2: Apply the quotient rule formula

\(f'(x)=\frac{(20x^{3}-70x)(x^{2}+6)-(5x^{4}-35x^{2}+30)(2x)}{(x^{2}+6)^{2}}\)
Expand the numerator:

$$
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Step3: Substitute \(x = 5\)

\(f'(5)=\frac{10\times5^{5}+120\times5^{3}-480\times5}{(5^{2}+6)^{2}}\)
Calculate the numerator:
\(10\times5^{5}=10\times3125 = 31250\), \(120\times5^{3}=120\times125=15000\), \(480\times5 = 2400\)
\(10\times5^{5}+120\times5^{3}-480\times5=31250+15000 - 2400=43850\)
Calculate the denominator: \((5^{2}+6)^{2}=(25 + 6)^{2}=31^{2}=961\)
\(f'(5)=\frac{43850}{961}\approx45.63\)

Answer:

\(\frac{43850}{961}\)