How do you find f'(x) using the definition of a derivative for f(x)=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)=9−x=uand u=9-xNow we have to derivate them:F′(u)=u12=12u−12Write the Expression as "pretty" as possibleand we get F′(u)=12⋅1u12=12⋅1uwe have to calculate u'u'=(9-x)'=-1The only ting left now is to fill in everything we have, into the formulaf′(x)=F′(u)⋅u′=12⋅1u⋅(−1)=−12⋅19−x

f(x)=9−xf′(x)=limh→0f(x+h)−f(x)h=limh→09−(x+h)−9−xh (Form 00)Rationalize the numerator.=limh→0(9−(x+h)−9−x)h⋅(9−(x+h))+9−x(9−(x+h))+9−x=limh→09−(x+h)−(9−x)h(9−(x+h))+9−x=limh→0−hh(9−(x+h))+9−x=limh→0−1h(9−(x+h))+9−x=−19−x+9−x=−129−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)=9−x=uand u=9-xNow we have to derivate them:F′(u)=u12=12u−12Write the Expression as "pretty" as possibleand we get F′(u)=12⋅1u12=12⋅1uwe have to calculate u'u'=(9-x)'=-1The only ting left now is to fill in everything we have, into the formulaf′(x)=F′(u)⋅u′=12⋅1u⋅(−1)=−12⋅19−x

f(x)=9−xf′(x)=limh→0f(x+h)−f(x)h=limh→09−(x+h)−9−xh (Form 00)Rationalize the numerator.=limh→0(9−(x+h)−9−x)h⋅(9−(x+h))+9−x(9−(x+h))+9−x=limh→09−(x+h)−(9−x)h(9−(x+h))+9−x=limh→0−hh(9−(x+h))+9−x=limh→0−1h(9−(x+h))+9−x=−19−x+9−x=−129−x

Get the solution to "How do you find f'(x) using the definition..."

Aliyah Mendez

Answered question

2022-04-15

How do you find f'(x) using the definition of a derivative for

f (x) = \sqrt{9 - x}

Answer & Explanation

xcentricccrhb1

Beginner2022-04-16Added 12 answers

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^{\frac{1}{2}} = \frac{1}{2} u^{- \frac{1}{2}}

Write the Expression as "pretty" as possible
and we get

F^{'} (u) = \frac{1}{2} \cdot \frac{1}{u^{\frac{1}{2}}} = \frac{1}{2} \cdot \frac{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) \cdot u^{'} = \frac{1}{2} \cdot \frac{1}{\sqrt{u}} \cdot (- 1) = - \frac{1}{2} \cdot \frac{1}{\sqrt{9 - x}}

StettyNagEragpouj

Beginner2022-04-17Added 7 answers

f (x) = \sqrt{9 - x}

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

= lim_{h \to 0} \frac{9 - (x + h) - \sqrt{9 - x}}{h}

(Form

\frac{0}{0}

)
Rationalize the numerator.

= lim_{h \to 0} \frac{(\sqrt{9 - (x + h) - \sqrt{9 - x}})}{h} \cdot \frac{(\sqrt{9 - (x + h)}) + \sqrt{9 - x}}{(\sqrt{9 - (x + h)}) + \sqrt{9 - x}}

= lim_{h \to 0} \frac{9 - (x + h) - (9 - x)}{h (\sqrt{9 - (x + h)}) + \sqrt{9 - x}}

= lim_{h \to 0} \frac{- h}{h (\sqrt{9 - (x + h)}) + \sqrt{9 - x}}

= lim_{h \to 0} \frac{- 1}{h (\sqrt{9 - (x + h)}) + \sqrt{9 - x}}

= \frac{- 1}{\sqrt{9 - x} + \sqrt{9 - x}}

= \frac{- 1}{2 \sqrt{9 - x}}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?