5.4 Indefinite Integrals and the Net Change Theorem/1: Difference between revisions
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<math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math> | <math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math> | ||
<br><br> | |||
<math>\begin{align} | <math>\begin{align} | ||
a &= x^2+1 & b &= a^{1/2} \\[0.6ex] | a &= x^2+1 & b &= a^{1/2} \\[0.6ex] | ||
\frac{da}{dx} &=2x y & \frac{db}{da} &=\frac{1}{2}a^{-1/2} | \frac{da}{dx} &=2x y & \frac{db}{da} &=\frac{1}{2}a^{-1/2} | ||
\end{align}</math> | \end{align}</math> | ||
<br><br> | |||
<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> | <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 17:15, 25 August 2022