Given the function $y =\sqrt x-1$, determine the slope of the tangent when x = 10. You must find the slope of the tangent using the method first principles.
I know how to find slope of a function but not at a specific point. If you could help me out that'll be great.
$\endgroup$ 21 Answer
$\begingroup$For the slope we obtain:$$\lim_{h\rightarrow0}\frac{\sqrt{10+h}-1-(\sqrt{10}-1)}{10+h-10}=\lim_{h\rightarrow0}\frac{\sqrt{10+h}-\sqrt{10}}{h}=$$$$=\lim_{h\rightarrow0}\frac{h}{h(\sqrt{10+h}+\sqrt{10})}=\frac{1}{2\sqrt{10}}.$$
$\endgroup$
