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

1 Answer
Mar 12, 2018

The slope is +- 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(x) is a compound function so it will be necessary to use the chain rule.

Denoting the two component functions as

g(x), where g(x) = sqrt(x)

and

h(x), where h(x) = x - 1

it might be noted that

f(x) = g(h(x)) (g of h of x)

The chain rule states

f'(x) = (g(h(x)))'

= g'(h(x))h'(x)

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(x) as

g(x) = (x)^(1/2)

so that the rules of polynomial differentiation might be easily applied

g'(x) = (1/2)(x)^(-1/2)

evaluating this at the value of the inner function

g'(h(x)) = (1/2)(x - 1)^(-1/2)

Noting

h'(x) = 1

The overall derivative is

f'(x) = (g(h(x)))' = g'(h(x))h'(x) = (1/2)(x - 1)^(-1/2)(1)

That is

f'(x) = 1/(2sqrt(x - 1))

so

f'(5) = 1/(2sqrt(5 - 1)) = 1/(2sqrt(4)) = +- 1/4