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

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}})