How to use the general for of the derivative formula with a fractional exponent

  • Thread starter mesa
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  • #1
mesa
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Homework Statement


f(x)=x^7/2, Find the derivative


Homework Equations



f'(a)=limit as x approaches a; ((x+h)-x)/h

The Attempt at a Solution


limit as h approached 0; ((x+h)^7/2-x^7/2)/h
Multiplied by the conjugate with -5/2 as exponent and came up with 2x^5/2, which is incorrect
 

Answers and Replies

  • #2
damabo
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do you want to evaluate [itex] \frac{x^7}{2}[/itex] or[itex] (x/2)^7 [/itex]?
in the first case, we get [itex] (\frac{(x+h)^7}{2} - \frac{x^7}{2})/h [/itex]. evaluating (x+h)^7 would require the binomial theorem, which says that [itex](x+y)^n = Ʃfrom k to n \frac{n!}{k!(n-k)!}x^{n-k}.y^k [/itex]
 
  • #3
Dick
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The conjugate of (x+h)^(7/2)-x^(7/2) is (x+h)^(7/2)+x^(7/2). Multiplying numerator and denominator by that is a good first step. Can you please show what you did?
 
  • #4
mesa
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The conjugate of (x+h)^(7/2)-x^(7/2) is (x+h)^(7/2)+x^(7/2). Multiplying numerator and denominator by that is a good first step. Can you please show what you did?

Okay we were multiplying it by (x+h)^(-5/2)+x^(-5/2), let me try it again
 
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  • #5
mesa
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Hmmm, okay this is a mess, how do deal with this mess (x+h)^7-x^7?
 
  • #6
Dick
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Hmmm, okay this is a mess, how do deal with this mess (x+h)^7-x^7?

Use the binomial theorem or just multiply it out. You will only wind up caring about the terms with highest power in x. Can you see why?
 
  • #7
mesa
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Im looking at it but dont see it
 
  • #8
Dick
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Im looking at it but dont see it

(x+h)^7=x^7+7*x^6*h+21*x^5*h^2+.... I'm using the binomial theorem. Why don't you care about terms that have h^2 or higher powers of h in them? Remember there's an h in the denominator and h will eventually go to zero.
 
  • #9
mesa
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I get that h goes to zero, but I thought we had to cancel it out first? Can we just assume zero for each term with h in it?
 
  • #10
Dick
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I get that h goes to zero, but I thought we had to cancel it out first? Can we just assume zero for each term with h in it?

No. Look, you really have to write out what you have gotten so far. Show your work.
 
  • #11
mesa
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Okay,

[(x+h)^7-x^7]/h((x+h)^(7/2)+x^7)=

[x^7+7x^6h+21x^5h^2+35x^4h^3+35x^3h^4+21x^2h^5+7xh^6+h^7-x^7]/h[(x+h)^(7/2)+x^(7/2)]
 
  • #12
Dick
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Okay,

[(x+h)^7-x^7]/h((x+h)^(7/2)+x^7)=

[x^7+7x^6h+21x^5h^2+35x^4h^3+35x^3h^4+21x^2h^5+7xh^6+h^7-x^7]/h[(x+h)^(7/2)+x^(7/2)]

Right. Thanks. You've got it. Cancel the x^7 in the numerator. Then divide the h in the denominator into the numerator. Now let h->0.
 
  • #13
mesa
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Right. Thanks. You've got it. Cancel the x^7 in the numerator. Then divide the h in the denominator into the numerator. Now let h->0.

Holy cow, that was a mess, ha ha!
It works out now. I was helping a friend with her calc and haven't done this in a little while. Looks like I need to review!

Thanks Dick!
 
  • #14
Dick
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Holy cow, that was a mess, ha ha!
It works out now. I was helping a friend with her calc and haven't done this in a little while. Looks like I need to review!

Thanks Dick!

Great! You see now that you only need to write (x+h)^7=x^7+7*x^6*h+... You don't need to care about the terms represented by the '...'. Right?
 
  • #15
mesa
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Great! You see now that you only need to write (x+h)^7=x^7+7*x^6*h+... You don't need to care about the terms represented by the '...'. Right?

Yup, got it, just like the others. First time doing the general form with fractional exponents, threw me through a loop, must be spending too much time in physics lol
 

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