question 11 · 1 point find the derivative of ( g(x)=7\tan^{-1}(x)) at the point ( x =-\frac{1}{3}). (enter…

question 11 · 1 point find the derivative of ( g(x)=7\tan^{-1}(x)) at the point ( x =-\frac{1}{3}). (enter an exact answer.) provide your answer below:

question 11 · 1 point find the derivative of ( g(x)=7\tan^{-1}(x)) at the point ( x =-\frac{1}{3}). (enter an exact answer.) provide your answer below:

Answer

Explanation:

Step1: Recall derivative formula

The derivative of $y = \tan^{- 1}(x)$ is $y'=\frac{1}{1 + x^{2}}$. For $g(x)=7\tan^{-1}(x)$, by the constant - multiple rule of differentiation $(cf(x))'=cf'(x)$, the derivative $g'(x)=\frac{7}{1 + x^{2}}$.

Step2: Evaluate at $x =-\frac{1}{3}$

Substitute $x =-\frac{1}{3}$ into $g'(x)$. We have $g'\left(-\frac{1}{3}\right)=\frac{7}{1+\left(-\frac{1}{3}\right)^{2}}$. First, calculate the denominator: $1+\left(-\frac{1}{3}\right)^{2}=1+\frac{1}{9}=\frac{9 + 1}{9}=\frac{10}{9}$. Then, $g'\left(-\frac{1}{3}\right)=\frac{7}{\frac{10}{9}}=7\times\frac{9}{10}=\frac{63}{10}$.

Answer:

$\frac{63}{10}$