Definition of the derivative to find the derivative of x^(1/3)


by Martinc31415
Tags: definition, derivative, x1 or 3
Martinc31415
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#1
Mar8-12, 11:09 AM
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1. The problem statement, all variables and given/known data

Use the definition of the derivative to find the derivative of x^(1/3)

2. Relevant equations



3. The attempt at a solution

[(x+h)^(1/3) - x^(1/3)]/h

I do not know where to go from here. If it were a square root I could conjugate.
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SammyS
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#2
Mar8-12, 11:26 AM
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Quote Quote by Martinc31415 View Post
1. The problem statement, all variables and given/known data

Use the definition of the derivative to find the derivative of x^(1/3)

2. Relevant equations

3. The attempt at a solution

[(x+h)^(1/3) - x^(1/3)]/h

I do not know where to go from here. If it were a square root I could conjugate.
Hello Martinc31415. Welcome ton PF !

The difference of cubes can be factored, [itex]a^3-b^3=(a-b)(a^2+ab+b^2)\,.[/itex]

So, suppose you have the difference of cube roots, [itex]\displaystyle P^{1/3}-Q^{1/3}[/itex]. In this case, [itex]\displaystyle P^{1/3} = a\,\ \text{ and }\ Q^{1/3} = b\,.[/itex]

Multiplying [itex]\displaystyle \left(P^{1/3}-Q^{1/3}\right)[/itex] by [itex]\displaystyle \left(P^{2/3}+P^{1/3}Q^{1/3}+Q^{2/3}\right)[/itex] will give [itex]\displaystyle \left(P^{1/3}\right)^3-\left(Q^{1/3}\right)^3=P-Q\,.[/itex]

Thus, [itex]\displaystyle \left(P^{2/3}+P^{1/3}Q^{1/3}+Q^{2/3}\right)[/itex] acts as the "conjugate" for [itex]\displaystyle \left(P^{1/3}-Q^{1/3}\right)\,.[/itex]
Martinc31415
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#3
Mar8-12, 01:42 PM
P: 2
oohh...

I would not have thought of that, ever!

Thanks a ton.


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