Processing Math: Done
Lösung 1.1:2e
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Lösning 1.1:2e moved to Solution 1.1:2e: Robot: moved page) |
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- | {{ | + | With the help of the square rule, we can expand the quadratic as |
- | < | + | |
- | {{ | + | |
+ | <math>\begin{align} | ||
+ | & f\left( x \right)=\left( x^{2}-1 \right)^{2}=\left( x^{2} \right)^{2}-2\centerdot x^{2}\centerdot 1+1^{2} \\ | ||
+ | & =x^{4}-2x^{2}+1 \\ | ||
+ | \end{align}</math> | ||
+ | |||
+ | |||
+ | When the function is written in this form, it is easy to differentiate term by term: | ||
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & {f}'\left( x \right)=\frac{d}{dx}\left( x^{4}-2x^{2}+1 \right) \\ | ||
+ | & =\frac{d}{dx}x^{4}-2\frac{d}{dx}x^{2}+\frac{d}{dx}1 \\ | ||
+ | & =4\centerdot x^{-1}-2\centerdot 2x^{-1}+0 \\ | ||
+ | & =4x^{3}-4x=4x\left( x^{2}-1 \right) \\ | ||
+ | \end{align}</math> |
Version vom 11:37, 10. Okt. 2008
With the help of the square rule, we can expand the quadratic as
x
=
x2−1
2=
x2
2−2
x2
1+12=x4−2x2+1
When the function is written in this form, it is easy to differentiate term by term:
x
=ddx
x4−2x2+1
=ddxx4−2ddxx2+ddx1=4
x−1−2
2x−1+0=4x3−4x=4x
x2−1