Question

if ( f(x)=6+\frac{5}{x}+\frac{6}{x^{2}} ), find ( f^{prime}(x) ).
find ( f^{prime}(4) ).

Explanation:

Step1: Rewrite the function

Rewrite \( f(x)=6 + \frac{5}{x}+\frac{6}{x^{2}} \) as \( f(x)=6 + 5x^{-1}+6x^{-2} \).

Step2: Apply the power rule

The power rule is \( \frac{d}{dx}(x^{n})=nx^{n - 1} \).
For the constant term \( 6 \), its derivative is \( 0 \) (since \( \frac{d}{dx}(c)=0 \), where \( c \) is a constant).
For the term \( 5x^{-1} \), using the power rule: \( \frac{d}{dx}(5x^{-1})=5\times(-1)x^{-1 - 1}=- 5x^{-2} \).
For the term \( 6x^{-2} \), using the power rule: \( \frac{d}{dx}(6x^{-2})=6\times(-2)x^{-2 - 1}=-12x^{-3} \).
So \( f^{\prime}(x)=-5x^{-2}-12x^{-3}=-\frac{5}{x^{2}}-\frac{12}{x^{3}} \).

Step3: Find \( f^{\prime}(4) \)

Substitute \( x = 4 \) into \( f^{\prime}(x) \).
\( f^{\prime}(4)=-\frac{5}{4^{2}}-\frac{12}{4^{3}} \).
First, calculate \( \frac{5}{4^{2}}=\frac{5}{16} \) and \( \frac{12}{4^{3}}=\frac{12}{64}=\frac{3}{16} \).
Then \( f^{\prime}(4)=-\frac{5}{16}-\frac{3}{16}=-\frac{5 + 3}{16}=-\frac{8}{16}=-\frac{1}{2} \).

Answer:

\( f^{\prime}(x)=-\frac{5}{x^{2}}-\frac{12}{x^{3}} \); \( f^{\prime}(4)=-\frac{1}{2} \)