Question

3.3 differentiation rules - ds2: problem 3
(6 points)
if
f(x)=\frac{6 x^{3}-5}{x^{4}}
find ( f^{prime}(x) ).
( f^{prime}(x)= )
find ( f^{prime}(3) ).
( f^{prime}(3)= )
note: you can earn partial credit on this problem.
note: you are in the reduced scoring period. all work counts for ( 85 % ) of the original.
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Explanation:

Step1: Simplify the function

Rewrite \( f(x)=\frac{6x^{3}-5}{x^{4}} \) as \( f(x)=6x^{- 1}-5x^{-4} \).

Step2: Differentiate using power rule

The power rule is \( \frac{d}{dx}(x^{n})=nx^{n - 1} \).
For \( y = 6x^{-1} \), \( y^\prime=6\times(-1)x^{-1 - 1}=-6x^{-2} \).
For \( y=-5x^{-4} \), \( y^\prime=-5\times(-4)x^{-4 - 1}=20x^{-5} \).
So \( f^\prime(x)=-6x^{-2}+20x^{-5}=\frac{-6}{x^{2}}+\frac{20}{x^{5}} \).

Step3: Find \( f^\prime(3) \)

Substitute \( x = 3 \) into \( f^\prime(x) \).
\( f^\prime(3)=\frac{-6}{3^{2}}+\frac{20}{3^{5}}=\frac{-6}{9}+\frac{20}{243} \).
First, \( \frac{-6}{9}=-\frac{2}{3}=-\frac{162}{243} \).
Then \( f^\prime(3)=-\frac{162}{243}+\frac{20}{243}=\frac{-162 + 20}{243}=\frac{-142}{243} \).

Answer:

\( f^\prime(x)=\frac{-6}{x^{2}}+\frac{20}{x^{5}} \), \( f^\prime(3)=\frac{-142}{243} \)