022 Exam 1 Sample A, Problem 1

1. Use the definition of derivative to find the derivative of $f(x)={\sqrt {x-5}}$.

Foundations:
Recall that the derivative is actually defined through the limit
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}.$
The goal in solving this is to plug the values $x+h$ and $x$ into the appropriate $f$ in the numerator, and then find a way to cancel the $h$ in the denominator. Unlike simplifying a rational expression containing radicals, here it's appropriate to have a radical in the denominator.
Again, the goal is to cancel the $h$.

 Solution:

Step 1:
Following the hints above, we initially have
$f'(x)\,=\,\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}\,=\,\lim _{h\rightarrow 0}{\frac {{\sqrt {x+h-5}}-{\sqrt {x-5}}}{h}}.$

Step 2:
We can't plug in $h=0$, as this would lead to the unallowed "division by zero". Instead, we multiply by the conjugate of the numerator and clean up:      $\lim _{h\rightarrow 0}{\frac {{\sqrt {x+h-5}}-{\sqrt {x-5}}}{h}}\cdot {\frac {{\sqrt {x+h-5}}+{\sqrt {x-5}}}{{\sqrt {x+h-5}}+{\sqrt {x-5}}}}$   $=\,\,\lim _{h\rightarrow 0}{\frac {\left({\sqrt {x+h-5}}\right)^{2}-\left({\sqrt {x-5}}\right)^{2}}{h\left({\sqrt {x+h-5}}+{\sqrt {x-5}}\right)}}$   $=\,\,\lim _{h\rightarrow 0}{\frac {x+h-5-\left(x-5\right)}{h\left({\sqrt {x+h-5}}+{\sqrt {x-5}}\right)}}$   $=\,\,\lim _{h\rightarrow 0}{\frac {h}{h\left({\sqrt {x+h-5}}+{\sqrt {x-5}}\right)}}$   $=\,\,\lim _{h\rightarrow 0}{\frac {1}{\left({\sqrt {x+h-5}}+{\sqrt {x-5}}\right)}}$   $=\,\,{\frac {1}{2{\sqrt {x-5}}}}.$
Notice that this is the same result you would get using our more convenient "rules of integration," including the chain rule, but that's not the point of this problem. You specifically need to treat the derivative as a limit.

Final Answer:
Note that no points would be given for only a correct answer. This exam question is about correctly using the definition of a derivative. However, the correct answer is
$f'(x)\,\,=\,\,{\frac {1}{2{\sqrt {x-5}}}}.$

Return to Sample Exam