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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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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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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  • phi
let'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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