5.4 Indefinite Integrals and the Net Change Theorem/1: Difference between revisions

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let <math>a=x^2+1</math> and <math>b=a^{1/2}</math> then <math>\frac{da}{dx}=2x \text{ and } \frac{db}{da}=\frac{1}{2}a^{-1/2}</math>
let <math>a=x^2+1</math> and <math>b=a^{1/2}</math> then <math>\frac{da}{dx}=2x \text{ and } \frac{db}{da}=\frac{1}{2}a^{-1/2}</math>


<math>\frac{da}{dx}\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>
<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>

Revision as of 16:30, 25 August 2022

let and then