Processing Math: Done
Lösung 1.2:3b
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Lösning 1.2:3b moved to Solution 1.2:3b: Robot: moved page) |
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- | {{ | + | The outer function in the expression is "the root of something", |
- | < | + | |
- | {{ | + | |
- | {{ | + | <math>\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}</math> |
- | + | ||
- | {{ | + | |
+ | and differentiating with the chain rule gives | ||
+ | |||
+ | |||
+ | <math>\frac{d}{dx}\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}=\frac{1}{2\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}}\centerdot \left( \frac{x+1}{x-1} \right)^{\prime }</math> | ||
+ | |||
+ | |||
+ | We establish the inner derivative by using the quotient rule, | ||
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & \frac{d}{dx}\sqrt{\frac{x+1}{x-1}}=\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{\left( x+1 \right)^{\prime }\centerdot \left( x-1 \right)-\left( x+1 \right)\centerdot \left( x-1 \right)^{\prime }}{\left( x-1 \right)^{2}} \\ | ||
+ | & =\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{1\centerdot \left( x-1 \right)-\left( x+1 \right)\centerdot 1}{\left( x-1 \right)^{2}} \\ | ||
+ | & =\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{-2}{\left( x-1 \right)^{2}} \\ | ||
+ | & =-\sqrt{\frac{x-1}{x+1}\centerdot }\frac{1}{\left( x-1 \right)^{2}} \\ | ||
+ | & =-\frac{1}{\left( x-1 \right)^{{3}/{2}\;}\sqrt{x+1}} \\ | ||
+ | \end{align}</math> | ||
+ | |||
+ | |||
+ | where we have used the simplification | ||
+ | |||
+ | |||
+ | <math>\frac{\sqrt{x-1}}{\left( x-1 \right)^{2}}=\frac{\left( x-1 \right)^{{1}/{2}\;}}{\left( x-1 \right)^{2}}=\left( x-1 \right)^{\frac{1}{2}-2}=\left( x-1 \right)^{-\frac{3}{2}}=\frac{1}{\left( x-1 \right)^{\frac{3}{2}}}</math> |
Version vom 12:45, 12. Okt. 2008
The outer function in the expression is "the root of something",
x−1x+1
and differentiating with the chain rule gives
x−1x+1
=12
x−1x+1
x−1x+1
We establish the inner derivative by using the quotient rule,
x−1x+1=12
x−1x+1
x−1
2
x+1
x−1
−
x+1
x−1
=12
x−1x+1
x−1
21
x−1
−
x+1
1=12
x−1x+1
−2
x−1
2=−
x+1x−1
1
x−1
2=−1
x−1
3
2
x+1
where we have used the simplification
x−1
x−1
2=
x−1
2
x−1
1
2=
x−1
21−2=
x−1
−23=1
x−1
23