Question

compute the derivative of the given function.
$j(r)=\tan^{-1}(8r)$

Explanation:

Step1: Recall the derivative formula for inverse tangent function

The derivative of \(y = \tan^{-1}(u)\) is \(y'=\frac{u'}{1 + u^{2}}\) by the chain rule. Here \(u = 8r\), and \(u'=\frac{d(8r)}{dr}=8\).

Step2: Substitute \(u\) and \(u'\) into the formula

Substitute \(u = 8r\) and \(u'=8\) into \(\frac{u'}{1 + u^{2}}\). We get \(j'(r)=\frac{8}{1+(8r)^{2}}\).

Step3: Simplify the denominator

Since \((8r)^{2}=64r^{2}\), then \(j'(r)=\frac{8}{1 + 64r^{2}}\).

Answer:

\(j'(r)=\frac{8}{1 + 64r^{2}}\)