Question

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

Answer 1

`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)`

Answer 2

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))`