1.1:2a alternativ 1

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(Die Seite wurde neu angelegt: <math>f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}</math> wobei <math>f(x)=x^2-3x+1</math> <math>\begin{align} f'(x)&=\lim_{h \to 0}\frac{(x+h)^{2}-3(x+h)+1-(x^{2}-3x+1}{h...)
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<math>f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}</math> wobei <math>f(x)=x^2-3x+1</math>
<math>f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}</math> wobei <math>f(x)=x^2-3x+1</math>
<math>\begin{align}
<math>\begin{align}
-
f'(x)&=\lim_{h \to 0}\frac{(x+h)^{2}-3(x+h)+1-(x^{2}-3x+1}{h}\\
+
f'(x)&=\lim_{h \to 0}\frac{(x+h)^{2}-3(x+h)+1-(x^{2}-3x+1)}{h}\\
&=\lim_{h \to 0}\frac{x^{2}+2hx+h^{2}-3x-3h+1-x^{2}+3x+1}{h}\\
&=\lim_{h \to 0}\frac{x^{2}+2hx+h^{2}-3x-3h+1-x^{2}+3x+1}{h}\\
&=\lim_{h \to 0}\frac{2hx+h^{2}-3h}{h}\\
&=\lim_{h \to 0}\frac{2hx+h^{2}-3h}{h}\\
&=\lim_{h \to 0}2x+h-3\\
&=\lim_{h \to 0}2x+h-3\\
&=2x-3\end{align}</math>
&=2x-3\end{align}</math>

Version vom 13:06, 5. Sep. 2009

\displaystyle f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} wobei \displaystyle f(x)=x^2-3x+1 \displaystyle \begin{align} f'(x)&=\lim_{h \to 0}\frac{(x+h)^{2}-3(x+h)+1-(x^{2}-3x+1)}{h}\\ &=\lim_{h \to 0}\frac{x^{2}+2hx+h^{2}-3x-3h+1-x^{2}+3x+1}{h}\\ &=\lim_{h \to 0}\frac{2hx+h^{2}-3h}{h}\\ &=\lim_{h \to 0}2x+h-3\\ &=2x-3\end{align}

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