Answer
$y=\displaystyle \frac{3}{5}x+\frac{16}{5}$
Work Step by Step
The slope of the tangent at x=a
is the derivative of f(x), calculated in x=a.
The tangent line has slope $f^{\prime}(a)$, and passes through $(a,f(a)).\\\\$
The point-slope equation is
$y-f(a)=f^{\prime}(a)(x-a)$
To find $f^{\prime}(x)$, we need the chain rule:
$f(x)=(x^{2}+16)^{1/2}=[w(x)]^{1/2}$
$f^{\prime}(x)=\displaystyle \frac{1}{2}[w(x)]^{-1/2}\cdot w^{\prime}(x)$
$f^{\prime}(x)=\displaystyle \frac{1}{2}(x^{2}+16)^{-1/2}\cdot 2x=\color{blue}{\frac{x}{\sqrt{x^{2}+16}}}$
$f(3)=\sqrt{9+16}=\sqrt{25}=5,$
$f^{\prime\prime}(3)=\displaystyle \frac{3}{\sqrt{9+16}}=\frac{3}{5}$
Point-slope form for the line passing through $(3,5)$, with slope $\displaystyle \frac{3}{5}$
$y-5=\displaystyle \frac{3}{5}(x-3)$
... solve for y (to transform to slope-intercept form)...
$y-5=\displaystyle \frac{3}{5}x-\frac{9}{5}$
$y=\displaystyle \frac{3}{5}x-\frac{9}{5}+5$
$y=\displaystyle \frac{3}{5}x+\frac{16}{5}$
