How do you find f'(x) using the definition of a derivative for f(x)=sqrt(9 − x)?

2 Answers
Nov 2, 2015

f'(x)=-1/(2sqrt(9-x))

Explanation:

The task is in the form f(x)=F(g(x))=F(u)
We have to use the Chain rule.

Chain rule: f'(x)=F'(u)*u'

We have F(u)=sqrt(9-x)=sqrt(u)
and u=9-x

Now we have to derivate them:

F'(u)=u^(1/2)'=1/2u^(-1/2)

Write the Expression as "pretty" as possible
and we get F'(u)=1/2*1/(u^(1/2))=1/2*1/sqrt(u)

we have to calculate u'
u'=(9-x)'=-1

The only ting left now is to fill in everything we have, into the formula

f'(x)=F'(u)*u'=1/2*1/sqrt(u)*(-1)=-1/2*1/sqrt(9-x)

Nov 3, 2015

To use the definition see the explanation section below.

Explanation:

f(x) = sqrt(9-x)

f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h

= lim_(hrarr0)(sqrt(9-(x+h)) - sqrt(9-x))/h (Form 0/0)

Rationalize the numerator.

= lim_(hrarr0)((sqrt(9-(x+h)) - sqrt(9-x)))/h * ((sqrt(9-(x+h)) + sqrt(9-x)))/((sqrt(9-(x+h)) + sqrt(9-x)))

= lim_(hrarr0)(9-(x+h)-(9-x))/(h (sqrt(9-(x+h)) + sqrt(9-x)))

= lim_(hrarr0)(-h)/(h (sqrt(9-(x+h)) + sqrt(9-x)))

= lim_(hrarr0)(-1)/ ((sqrt(9-(x+h)) + sqrt(9-x))

= (-1)/(sqrt(9-x)+sqrt(9-x)

= (-1)/(2sqrt(9-x))