# How do you find the slope of the curve f(x)=sqrt(x-1) at the point x=5?

Mar 12, 2018

The slope is $\pm \frac{1}{4}$
(there are two possible values as the function is not one-one as it contains a square root)

#### Explanation:

The slope of a function at some particular value of the independent variable is the first derivative evaluated at that particular value.

$f \left(x\right)$ is a compound function so it will be necessary to use the chain rule.

Denoting the two component functions as

$g \left(x\right)$, where $g \left(x\right) = \sqrt{x}$

and

$h \left(x\right)$, where $h \left(x\right) = x - 1$

it might be noted that

$f \left(x\right) = g \left(h \left(x\right)\right)$ ($g$ of $h$ of $x$)

The chain rule states

$f ' \left(x\right) = \left(g \left(h \left(x\right)\right)\right) '$

$= g ' \left(h \left(x\right)\right) h ' \left(x\right)$

That is, the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

It is convenient to write $g \left(x\right)$ as

$g \left(x\right) = {\left(x\right)}^{\frac{1}{2}}$

so that the rules of polynomial differentiation might be easily applied

$g ' \left(x\right) = \left(\frac{1}{2}\right) {\left(x\right)}^{- \frac{1}{2}}$

evaluating this at the value of the inner function

$g ' \left(h \left(x\right)\right) = \left(\frac{1}{2}\right) {\left(x - 1\right)}^{- \frac{1}{2}}$

Noting

$h ' \left(x\right) = 1$

The overall derivative is

$f ' \left(x\right) = \left(g \left(h \left(x\right)\right)\right) ' = g ' \left(h \left(x\right)\right) h ' \left(x\right) = \left(\frac{1}{2}\right) {\left(x - 1\right)}^{- \frac{1}{2}} \left(1\right)$

That is

$f ' \left(x\right) = \frac{1}{2 \sqrt{x - 1}}$

so

$f ' \left(5\right) = \frac{1}{2 \sqrt{5 - 1}} = \frac{1}{2 \sqrt{4}} = \pm \frac{1}{4}$