Question

solve the one in the middle asking for f'(10)​

Answer 1

From the definition, we have

$\displaystyle f'(10) = \lim_{x\to0} \frac{f(x) - f(10)}{x - 10}$

With $f(x)=\sqrt{x-1}$, we get $f(10)=\sqrt{10-1}=\sqrt9=3$. So the limit we want to compute is

$\displaystyle f'(10) = \lim_{x\to0} \frac{\sqrt{x-1} - 3}{x - 10}$

Rationalize the numerator by multiplying by its conjugate:

$\displaystyle \frac{\sqrt{x-1} - 3}{x - 10} \times \frac{\sqrt{x-1}+3}{\sqrt{x-1}+3} = \frac{\left(\sqrt{x-1}\right)^2-3^2}{(x-10)\left(\sqrt{x-1}+3\right)} = \frac{x-10}{(x-10)\left(\sqrt{x-1}+3\right)}$

x is approaching 10, which is to say x ≠ 10, so we can cancel the factors of x - 10 and remove the discontinuity.

Then we're left with

$\displaystyle f'(10) = \lim_{x\to0} \frac1{\sqrt{x-1}+3} = \frac1{\sqrt{10-1}+3} = \frac1{\sqrt9+3} = \frac1{3+3} = \boxed{\frac16}$

evaluated by direct substitution, which we can do since the limand is continuous.