Answer
$ f'(4)=\frac{15}{16}$
Equation of the tangent line is:
$ y=\frac{15}{16}x+\frac{1}{2}$
Work Step by Step
Recall that $ f'(a)=\lim\limits_{x \to a}\frac{f(x)-f(a)}{x-a}$
Given that $ a=4$ and $ f(x)=x+x^{-1}$. Therefore, $ f'(4)=\lim\limits_{x \to 4}\frac{f(x)-f(4)}{x-4}=\lim\limits_{x \to 4}\frac{(x+\frac{1}{x})-(4+\frac{1}{4})}{x-4}$
$=\lim\limits_{x \to 4}\frac{\frac{x^{2}+1}{x}-\frac{4^{2}+1}{4}}{x-4}=\lim\limits_{x \to 4}\frac{4x^{2}+4-16x-x}{4x(x-4)}=\lim\limits_{x \to 4}\frac{4x^{2}-17x+4}{4x(x-4)}=\lim\limits_{x \to 4}\frac{(4x-1)(x-4)}{4x(x-4)}=\lim\limits_{x \to 4}\frac{4x-1}{4x}=\frac{15}{16}$
Equation of the tangent line is of the form
$ y-f(a)=f'(a)(x-a)$
Knowing that $ f(a)=f(4)=\frac{17}{4}, f'(a)=f'(4)=\frac{15}{16}$ and $ a=4$, we obtain the equation of the tangent line through $ a $ as below.
$ y-\frac{17}{4}=\frac{15}{16}(x-4)$
Or $ y=\frac{15}{16}x+\frac{1}{2}$
