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Derivative using the definition

  • Thread starter Flappy
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  • #1
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Homework Statement



Use the definition of the derivative to show f is differentiable at x = 2.

f(x) = 1 + cos(pix/2)

The Attempt at a Solution



I started getting the solution but got stuck at one part.

f`(x) = lim h->0 | [ f(x+h) - f(x) ] / (h)

= lim h->0 | [f(2+h) - f(2) ] / (h)

= lim h -> 0 | [1 + cos(pi(2+h)/2) - 0 ] / (h)

= lim h -> 0 | [1 + cos(2pi+hpi)/2 ] / (h)

I know im supposed to get the h's to cancel out but im having trouble getting there
 
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Answers and Replies

  • #2
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[tex]f(x)=1+\cos\left(\frac{\pi x}{2}\right)[/tex]

[tex]\lim_{h\rightarrow0}=\frac{f(2+h)-f(2)}{h}[/tex]

[tex]\lim_{h\rightarrow0}=\frac{1+\cos\left[\frac{\pi}{2}(2+h)\right]-(1+\cos \pi)}{h}[/tex]

[tex]\lim_{h\rightarrow0}=\frac{1+\cos\left({\pi+\frac{\pi h}{2}}\right)}{h}[/tex]

[tex]\cos{(x+y)}=\cos x\cos y-\sin x\sin y[/tex]

[tex]\lim_{h\rightarrow0}=\frac{1-cos\left(\frac{\pi h}{2}\right)}{h}\cdot\frac{1+cos\left(\frac{\pi h}{2}\right)}{1+cos\left(\frac{\pi h}{2}\right)}[/tex]

*Difference of Squares

[tex]\lim_{h\rightarrow0}=\frac{\sin{\left(\frac{\pi h}{2}\right)}}{h}\cdot\frac{\sin{\left(\frac{\pi h}{2}\right)}}{1+cos\left(\frac{\pi h}{2}\right)}\cdot\frac{\frac{\pi}{2}}{\frac{\pi}{2}}[/tex]

Still typing, comp freezes! Refresh every minute if you want.
 
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  • #3
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Is it, pi/2 or just pi? Reading your last 2 lines, there is a discrepancy.
 
  • #4
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Oh, I might have mislead with the argument inside cos. It's

[tex]f(x)=1+\cos\left(\frac{\pi(x)}{2}\right)[/tex]
 
  • #5
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So what I have so far is good?
 
  • #6
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Yeah, I see what you did. I was a little confused at first. I forgot to add the 2 under the division on my last 2 lines. Thank you for taking the time by the way.
 
  • #7
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Hey hold on, my friend needs help with graphing (she's a girl, so :p) But hmm ... basically my LaTeX image isn't showing up. What you want to do is, use a trig identity ... 1-cos^2 x = sin^2 x. Use the fact that sinx\x = 1. So your angle is pi*h/2, so what will you need to do? Think back! I think it will work, I haven't finished working the problem. Been having to type the LaTeX, lol.
 
  • #8
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Dam, this problem is just nasty.

Practicing latex here. this is after you multiply by the conjugate right?

[tex]\lim_{h\rightarrow0}= \frac{1 - cos(\frac{{\pi}h}{2})^2}{h(1+cos(\frac{{\pi}h}{2})} [/tex]
 
  • #9
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Lol, yes. Damn I can't get it to work. There is an easier way. You just gotta be neat about it yanno. If you want to make a complex fraction. Put ...

\frac{a}{b}

Then to the side

a=\frac{c}{d}
b=\frac{e}{f}

Then just replace by copy paste :p But obviously I haven't learned my lesson ... argh. And you don't need the brackets if it's just one letter.

\frac a b works great!
 
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  • #10
1,752
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I don't know if I'm on the right track with the problem b/c when I differentiate it doesn't match the definition. I will get 0 by using the definition ... hmm. But of course, we evaluated at 2!!!
 
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  • #11
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WOOHOO!!! It's right, lol. Forgot we evaluated at 2 :p

I was like ... wtf I typed all that up for nothing! Damn I suck.
 
  • #12
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I believe you are supposed to get 0. Unless im not following you. I did the derivative using rules and ended up getting 0.

Doing the derivative I got [tex]-sin\frac{{\pi}x}{2} * \frac{{\pi}}{2}[/tex]

Then by plugging in 2 you would get 0 * pi/2 which is 0.
 
  • #13
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I'm done typing it up, whew! Btw, there should be a negative somewhere ... not sure where it went, since the derivative of cosx is -sinx. It would be counted wrong even though we get the right answer yanno. Just letting you know. Maybe it disappeared since we evaluated at 2 from the get go :p
 
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  • #14
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Just a question, where did the pi/2 multiplication come from?
 
  • #15
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Just a question, where did the pi/2 multiplication come from?
What is the limit of ... [tex]\lim_{h\rightarrow0}\frac{\sin{\left(\frac{\pi h}{2}\right)}}{h}[/tex] ? Definitely not 1 since it's not [tex]\lim_{h\rightarrow0}\frac{\sin x}{x}=1 [/tex]

So I have to multiply both numerator and denominator by pi/2, which is implied multiplication of 1. Basically, the angle and the denominator must be the same!
 
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  • #16
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I see, i see. Well, thank you for all the help, appreciated.
 
  • #17
1,752
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I see, i see. Well, thank you for all the help, appreciated.
Anytime :) If you want help like this again, make sure to show work :-] Only reason why I tried my best since I know you tried! Have fun, Calculus gets better.
 

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