University Calculus: Early Transcendentals (3rd Edition)

The slope of the tangent line to the curve at $x=4$ is $-1/16$.
$$y=f(x)=\frac{1}{\sqrt x}$$ The slope of the tangent line to the curve at $x=4$ is $$y'=\lim_{h\to0}\frac{f(4+h)-f(4)}{h}=\lim_{h\to0}\frac{\frac{1}{\sqrt{h+4}}-\frac{1}{\sqrt 4}}{h}=\lim_{h\to0}\frac{\frac{1}{\sqrt{h+4}}-\frac{1}{2}}{h}$$ $$y'=\lim_{h\to0}\frac{2-\sqrt{h+4}}{2h\sqrt{h+4}}$$ Multiply both numerator and denominator by $2+\sqrt{h+4}$: $$y'=\lim_{h\to0}\frac{(2-\sqrt{h+4})(2+\sqrt{h+4})}{2h\sqrt{h+4}(2+\sqrt{h+4})}=\lim_{h\to0}\frac{4-(h+4)}{2h\sqrt{h+4}(2+\sqrt{h+4})}$$ $$y'=\lim_{h\to0}\frac{-h}{2h\sqrt{h+4}(2+\sqrt{h+4})}=\lim_{h\to0}\frac{-1}{2\sqrt{h+4}(2+\sqrt{h+4})}$$ $$y'=\frac{-1}{2\sqrt{0+4}(2+\sqrt{0+4})}=\frac{-1}{2\sqrt4(2+\sqrt4)}$$ $$y'=\frac{-1}{2\times2\times4}=-\frac{1}{16}$$