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

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Answer 1 · 2018-03-12T20:58:41

The slope is #+- 1/4#
(there are two possible values as the function is not one-one as it contains a square root)

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#