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anonymous
 2 years ago
I'm working is session 2 of Differentiation, and I'm clueless how these numbers came out for solving the secant. [(x0)(x0+Δx)/(x0)(x0+Δx)] * [(1\x0+Δx  1/x0) / Δx] when the previous equation was just the equation after the multiplication symbol?
anonymous
 2 years ago
I'm working is session 2 of Differentiation, and I'm clueless how these numbers came out for solving the secant. [(x0)(x0+Δx)/(x0)(x0+Δx)] * [(1\x0+Δx  1/x0) / Δx] when the previous equation was just the equation after the multiplication symbol?

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phi
 2 years ago
Best ResponseYou've already chosen the best response.0let's use A and B (easier to type) you start with \[ \frac{1}{A} + \frac{1}{B}\] to add the fractions you need a common denominator AB you can multiply the first fraction by B/B and the second by A/A \[ \frac{1}{A} \cdot \frac{B}{B} + \frac{1}{B} \cdot \frac{A}{A} = \frac{B}{AB}+\frac{A}{AB}= \frac{B+A}{AB}\] another way to do this is to multiply the fractions by AB/AB \[ \frac{AB}{AB} \left( \frac{1}{A} + \frac{1}{B}\right) \] if we distribute we get \[ \frac{1}{A} \cdot \frac{AB}{AB}+ \frac{1}{B}\cdot \frac{AB}{AB} \] which gives \[ \frac{B}{AB}+ \frac{A}{AB} \] The idea of multiplying by AB/AB (clearing the denominator) is just a way to add the fractions
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