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