Question

question find the derivative of $g(x)=5cot^{- 1}(x)$ at the point $x = \frac{7}{3}$. (enter an exact answer.) provide your answer below: $g(\frac{7}{3})=$

Explanation:

Step1: Recall derivative formula

The derivative of $y = \cot^{- 1}(x)$ is $y'=-\frac{1}{1 + x^{2}}$. Using the constant - multiple rule, if $g(x)=5\cot^{-1}(x)$, then $g'(x)=5\times(-\frac{1}{1 + x^{2}})=-\frac{5}{1 + x^{2}}$.

Step2: Substitute $x = \frac{7}{3}$

Substitute $x=\frac{7}{3}$ into $g'(x)$. We get $g'(\frac{7}{3})=-\frac{5}{1+(\frac{7}{3})^{2}}$.
First, calculate $1 + (\frac{7}{3})^{2}=1+\frac{49}{9}=\frac{9 + 49}{9}=\frac{58}{9}$.
Then $g'(\frac{7}{3})=-\frac{5}{\frac{58}{9}}=-5\times\frac{9}{58}=-\frac{45}{58}$.

Answer:

$-\frac{45}{58}$