Question

determine each of the following, where $f(x)=4+\frac{3}{x}+\frac{7}{x^{2}}$.
a. $f(x)=$
b. $f(2)=$

Explanation:

Step1: Rewrite function

Rewrite $f(x)=4 + \frac{3}{x}+\frac{7}{x^{2}}$ as $f(x)=4 + 3x^{-1}+7x^{-2}$.

Step2: Apply power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. For $f(x)=4 + 3x^{-1}+7x^{-2}$, the derivative of a constant 4 is 0, the derivative of $3x^{-1}$ is $3\times(-1)x^{-1 - 1}=-3x^{-2}$, and the derivative of $7x^{-2}$ is $7\times(-2)x^{-2 - 1}=-14x^{-3}$. So, $f^\prime(x)=- \frac{3}{x^{2}}-\frac{14}{x^{3}}$.

Step3: Evaluate $f^\prime(2)$

Substitute $x = 2$ into $f^\prime(x)$. $f^\prime(2)=-\frac{3}{2^{2}}-\frac{14}{2^{3}}=-\frac{3}{4}-\frac{14}{8}=-\frac{3}{4}-\frac{7}{4}=-\frac{3 + 7}{4}=-\frac{10}{4}=-\frac{5}{2}$.

Answer:

a. $f^\prime(x)=-\frac{3}{x^{2}}-\frac{14}{x^{3}}$
b. $f^\prime(2)=-\frac{5}{2}$