for y=1/(sqrt(x)) find slope of tangent line at the point where x=a
then, find the equation of the tangent line at (4,1/2)
The slope of the tangent line at $\displaystyle x=a$ is just $\displaystyle \left.\frac{\,dy}{\,dx}\right|_{x=a}$
Thus, $\displaystyle y=x^{-\frac{1}{2}}\implies \frac{\,dy}{\,dx}=-\tfrac{1}{2}x^{-\frac{3}{2}}$.
Thus, at $\displaystyle x=a$, $\displaystyle \left.\frac{\,dy}{\,dx}\right|_{x=a}=-\tfrac{1}{2}a^{-\frac{3}{2}}$
Now, let's find the equation of the tangent line:
$\displaystyle y-f(a)=f'(a)(x-a)$
Since $\displaystyle a=4,~f(a)=\tfrac{1}{2},~\text{and }f'(a)=-\tfrac{1}{16}$, we now see that $\displaystyle y=-\tfrac{1}{16}(x-4)+\tfrac{1}{2}\implies {\color{red}y=-\tfrac{1}{16}x+\tfrac{3}{4}}$
Does this make sense?
--Chris