Using the limit definition (f(x+h)-f(x)\h

  • #1
I have to differentiate some problems using the limit definition (f(x+h)-f(x)\h), but Iam having some trouble on a couple.

1. Square root of 30 is 0 because its a constant correct?

2. R(t) = 5t^(-3/5) <---- I have tried this problem many times with the definition, but I cant come up with the correct solution and I dont know why!

Thanks
 

Answers and Replies

  • #2
1,235
1
1. correct


2. rewrite this as [tex] 5\frac{d}{dt} \frac{1}{t^{\frac{3}{5}}} [/tex].

So: [tex] \frac{1}{\sqrt[5]{(t+h)^{3}}} - \frac{1}{\sqrt[5]{t^{3}}} [/tex]. Subtract the fractions. Then rationalize the numerator. take the limit, and then multiply by 5.
 
Last edited:
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
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1. Wow, I hate it when people misuse terminology so badly!!

NO!!! [itex]\sqrt{30}[/itex] is NOT 0!!!. It is about 5.4! Of course, it is true that the derivative of the function [itex]f(x)= \sqrt{30}[/itex], since that is a constant function, is 0.
 
  • #4
hmm iam still confused.. sorry
not about square root of 30.. about the next problem
 
  • #5
2,076
2
courtigrad, you surely meant the derivative wrt. t, didn't you? ;)
 
  • #6
1,235
1
yeah, just a typo. my bad. also i notice people type "courtigrad" instead of "courtrigrad".
 
  • #7
1,235
1
the trouble is: what do you multiply the fraction by to rationalize the numerator.

[tex] t^{\frac{3}{5}} - (t+h)^{\frac{3}{5}} [/tex] multiplied by something to get [tex] t - (t+h) = -h [/tex]

For example, [tex] t^{\frac{1}{2}} - (t+h)^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}}) = -h [/tex]
 
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  • #8
Alright I got it.. Thanks Guys!
 

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