Solving a Derivative Problem Using the Power Rule

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Homework Help Overview

The discussion revolves around a derivative problem involving polynomial and rational functions, specifically using the power rule for differentiation. The original poster expresses confusion regarding their calculations and the discrepancies with the book's answer.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conversion of terms into exponent form, particularly focusing on the correct application of negative exponents. There are questions about specific transformations, such as converting \(\frac{7}{T}\) and \(\frac{1}{2T}\) into their respective exponent forms.

Discussion Status

The discussion is ongoing, with participants providing clarifications and corrections regarding exponent rules. There is recognition of the need for a solid understanding of algebraic principles before tackling calculus problems. No explicit consensus has been reached on the original poster's derivative calculations.

Contextual Notes

Participants are navigating through potential misunderstandings of exponent rules and their implications for differentiation. The original poster's attempts at solving the derivative problem are met with questions about their algebraic manipulations.

swears
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I tried this derivative problem, but the back of the book shows a different answer then what I got. Can someone explain what I'm doing wrong.

[tex]3T^5 -5T^.5 + \frac{7}{T}[/tex]

So I did this:

[tex]15T^4 - 2.5T^-.5 - 7T^-8[/tex]

It's the last part I got wrong I'm not sure why. I converted 7/T to T^-7. Is that right?
 
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swears said:
It's the last part I got wrong I'm not sure why. I converted 7/T to T^-7. Is that right?

Not quite;

[tex]\frac{7}{T} = 7T^{-1}[/tex]
 
hmm, it's not doing the exponents right, u might have too look at the code.
 
Hootenanny said:
Not quite;

[tex]\frac{7}{T} = 7T^{-1}[/tex]

Oh, does the -1 come from the 7 or T?
 
swears said:
Oh, does the -1 come from the 7 or T?
It comes from the T, remember;

[tex]\frac{1}{T} = T^{-1}[/tex]
 
So [tex]\frac {1}{2T} = 1T^{-2}[/tex]?
 
See below message.
Hootenanny said:
It comes from the T, remember;

[tex]\frac{1}{T} = T^{-1}[/tex]
 
swears said:
So [tex]\frac {1}{2T} = 1T^{-2}[/tex]?

Nope, not quite.

[tex]\frac{1}{2T} = (2T)^{-1}[/tex]

The power applies to all terms of the denominator. However;

[tex]\frac{1}{T^2} = T^{-2}[/tex]

Does that make sense?
 
So, you basically just combine the top and bottom terms and negate the exponent.
 
  • #10
swears said:
So, you basically just combine the top and bottom terms and negate the exponent.
The exponent is the most important part. You can't just negate it.
 
  • #11
Ok, well I had this other problem Maybe I can get it right here.

[tex]\frac{t^2 + t^3 -{1}}{t^4}[/tex]

I divided by T^4 and then used the power rule and I got: [tex].5T^{-.5} + .75T^{-.25} + 4T^{-5}[/tex]
 
Last edited:
  • #12
I think you need to master algebra before trying calculus.

See: http://www.allaboutcircuits.com/vol_5/chpt_4/5.html
 
  • #13
Thanks, I got it.
 

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