Derivative help

  • Thread starter ussjt
  • Start date
41
0
If https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image1.png [Broken] , find [PLAIN]https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image2.png [Broken] [Broken]using the definition of derivative. [PLAIN]https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image2.png [Broken] [Broken] is the limit as https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image4.png [Broken] of the expression (answer)
I need help with finding the expression.

The value of the limit is 0.

Any help would be great.
 
Last edited by a moderator:

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,828
14
What have you done?
 

VietDao29

Homework Helper
1,415
1
f'(x0) is defined to be
[tex]f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}[/tex]
Apply that to your function:
[tex]f'(9) = \lim_{h \rightarrow 0} \frac{f(9 + h) - f(9)}{h}[/tex]. Now, how can you do this limit?
 
41
0
I am just confused with the 1/2^.5 part.

I am used to seeing something like f(x)= 1/x....so that you make it 1/(9+h).
 

VietDao29

Homework Helper
1,415
1
[tex]f(x) := \frac{1}{\sqrt{2}}[/tex] means that this function is a constant and does not depend on what x you choose.
It will return the same value (i.e, 1 / (2.5)), no matter what x is. For example:
If [tex]g(x) := \frac{1}{\sqrt{3}}[/tex], then:
[tex]g(0) = \frac{1}{\sqrt{3}}[/tex],
[tex]g(7) = \frac{1}{\sqrt{3}}[/tex],
[tex]g(9) = \frac{1}{\sqrt{3}}[/tex],...
Do you understand? Can you solve the problem now?
 
786
0
And if the limit definition doesn't click, look at it graphically. A constant function is just a horizontal line, which obviously has a slope of 0.
 
41
0
I have looked at it graphically, but I having a hard time understanding how to make it into an expression...lim h->0 something.

thanks for all of the input, but I am still confused.
 
786
0
Ok. VietDao29 showed you the limit definition of a derivative. It's the following...

[tex]f'(x)=\lim_{h\rightarrow{0}} \frac{f(x+h)-f(x)}{h}[/tex]

Your function is [tex]\frac{1}{\sqrt{2}}[/tex], so I'll start you out.

[tex]f'(9)=\lim_{h\rightarrow{0}} \frac{f(9+h)-f(9)}{h}[/tex]

This is still from VietDao29's post.

What is f(9)? What is f(0)? What is f of anything? If you can answer that, you can do your limit.
 

VietDao29

Homework Helper
1,415
1
Another hint is that:
[tex]\frac{f(9 + h) - f(9)}{h} = \frac{\frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}}{h} = 0, \ \forall h \neq 0[/tex]
What does the above expression tell you?
 
41
0
thanks, I got it now. Thanks a lot!
 

Want to reply to this thread?

"Derivative help" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top