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

2 Answers
Monica
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)#

Jim H
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))#

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