5.4 Indefinite Integrals and the Net Change Theorem/1: Difference between revisions
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<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 \frac{db}{da} &=\frac{1}{2}a^{-1/2} | \frac{da}{dx} &=2x & \frac{db}{da} &=\frac{1}{2}a^{-1/2} | ||
\end{align}</math> | \end{align}</math> | ||
<br><br> | <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> | ||
Latest revision as of 18:24, 26 August 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int\frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+c}
Show that: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} a &= x^2+1 & b &= a^{1/2} \\[0.6ex] \frac{da}{dx} &=2x & \frac{db}{da} &=\frac{1}{2}a^{-1/2} \end{align}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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}}}