if $f(x)=7 + \\frac{3}{x}+\\frac{5}{x^{2}}$, find $f(x)$. find $f(5)$.
if $f(x)=7 + \\frac{3}{x}+\\frac{5}{x^{2}}$, find $f(x)$. find $f(5)$.
Answer
Explanation:
Step1: Recall power - rule for differentiation
The power - rule states that if $y = x^n$, then $y^\prime=nx^{n - 1}$, and the derivative of a constant is 0. Given $f(x)=7+\frac{3}{x}+\frac{5}{x^{2}}=7 + 3x^{-1}+5x^{-2}$. The derivative of the constant 7 is 0, the derivative of $3x^{-1}$ is $3\times(-1)x^{-1 - 1}=-3x^{-2}$, and the derivative of $5x^{-2}$ is $5\times(-2)x^{-2 - 1}=-10x^{-3}$. So $f^\prime(x)=-3x^{-2}-10x^{-3}$.
Step2: Evaluate $f^\prime(5)$
Substitute $x = 5$ into $f^\prime(x)$. $f^\prime(5)=-3\times5^{-2}-10\times5^{-3}$. First, $5^{-2}=\frac{1}{5^{2}}=\frac{1}{25}$ and $5^{-3}=\frac{1}{5^{3}}=\frac{1}{125}$. So $f^\prime(5)=-3\times\frac{1}{25}-10\times\frac{1}{125}$. $-3\times\frac{1}{25}=-\frac{3}{25}=-\frac{15}{125}$, and $-10\times\frac{1}{125}=-\frac{10}{125}$. Then $f^\prime(5)=-\frac{15}{125}-\frac{10}{125}=-\frac{15 + 10}{125}=-\frac{25}{125}=-\frac{1}{5}$.
Answer:
$-\frac{1}{5}$