Question

3.3 differentiation rules - ds
(6 points)
let ( f ( x ) = \frac { 3 x ^ { 2 } + 3 x + 7 } { sqrt { x } } ). find the following:
( f ^ { prime } ( x ) = )
( f ^ { prime } ( 9 ) = )
note: you can earn partial credit on this problem.
note: you are in the reduced scoring period. all work counts for 85%
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Explanation:

Step1: Simplify the function

Rewrite \( f(x)=\frac{3x^{2}+3x + 7}{\sqrt{x}}=3x^{2-\frac{1}{2}}+3x^{1-\frac{1}{2}}+7x^{-\frac{1}{2}}=3x^{\frac{3}{2}}+3x^{\frac{1}{2}}+7x^{-\frac{1}{2}}\)

Step2: Apply the power rule \( (x^{n})^\prime=nx^{n - 1}\)

\( f^\prime(x)=3\times\frac{3}{2}x^{\frac{3}{2}-1}+3\times\frac{1}{2}x^{\frac{1}{2}-1}+7\times(-\frac{1}{2})x^{-\frac{1}{2}-1}\)
\( f^\prime(x)=\frac{9}{2}x^{\frac{1}{2}}+\frac{3}{2}x^{-\frac{1}{2}}-\frac{7}{2}x^{-\frac{3}{2}}\)

Step3: Calculate \( f^\prime(9)\)

Substitute \( x = 9\) into \( f^\prime(x)\):
\( f^\prime(9)=\frac{9}{2}\sqrt{9}+\frac{3}{2}\frac{1}{\sqrt{9}}-\frac{7}{2}\frac{1}{(\sqrt{9})^{3}}\)
\( f^\prime(9)=\frac{9}{2}\times3+\frac{3}{2}\times\frac{1}{3}-\frac{7}{2}\times\frac{1}{27}\)
\( f^\prime(9)=\frac{27}{2}+\frac{1}{2}-\frac{7}{54}\)
\( f^\prime(9)=\frac{27 + 1}{2}-\frac{7}{54}=\frac{28}{2}-\frac{7}{54}=14-\frac{7}{54}=\frac{14\times54-7}{54}=\frac{756 - 7}{54}=\frac{749}{54}\)

Answer:

\( f^\prime(x)=\frac{9}{2}x^{\frac{1}{2}}+\frac{3}{2}x^{-\frac{1}{2}}-\frac{7}{2}x^{-\frac{3}{2}}\), \( f^\prime(9)=\frac{749}{54}\)