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These are a bit more involved than exponents with e as a base, |
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| but not THAT much. | ||
| The derivative of 2 X is 2 X ln X. | ||
| The general rule is: | ||
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| Here's why, start with: | ||
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Y = aX |
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| Take the log (ln that is) of both sides: | ||
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ln Y = ln aX |
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| Use the log with exponents trick: | ||
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ln Y = X ln a |
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| Now solve for X: | ||
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| ln a is a constant. ln Y is a variable. | ||
| So it's easier to see how this will work, we will separate the two. | ||
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| Now we take the derivative of X with respect to Y. | ||
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| WHAT? Where did that 1/Y come from? | ||
| Don't you remember the last page? | ||
| The derivative of ln Y is 1/Y. | ||
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| Our original problem was Y = aX, so substitute that in for Y ... | ||
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| But we wanted dY/dX not dX/dY! | ||
| OK, so flip it over. | ||
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| The original equation was Y = aX. That means that if: | ||
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| And there it is! | ||
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copyright 2005 Bruce Kirkpatrick |
