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Definition of the derivative to find the derivative of x^(1/3) |
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| Mar8-12, 11:09 AM | #1 |
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Definition of the derivative to find the derivative of x^(1/3)
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. |
| Mar8-12, 11:26 AM | #2 |
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Mentor
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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] |
| Mar8-12, 01:42 PM | #3 |
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oohh...
I would not have thought of that, ever! Thanks a ton. |
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