Lösung 1.1:2e

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We expand the quadratic expression as
We expand the quadratic expression as
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
f(x) &= \bigl(x^2-1\bigr)^2\\[5pt]
f(x) &= \bigl(x^2-1\bigr)^2\\[5pt]
&= \bigl(x^2\bigr)^2 - 2\cdot x^2\cdot 1 + 1^2\\[5pt]
&= \bigl(x^2\bigr)^2 - 2\cdot x^2\cdot 1 + 1^2\\[5pt]
Zeile 9: Zeile 9:
When the function is written in this form, it is easy to differentiate term by term,
When the function is written in this form, it is easy to differentiate term by term,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^4-2x^2+1\bigr)\\[5pt]
f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^4-2x^2+1\bigr)\\[5pt]
&= \frac{d}{dx}\,x^4 - 2\frac{d}{dx}\,x^2 + \frac{d}{dx}\,1\\[5pt]
&= \frac{d}{dx}\,x^4 - 2\frac{d}{dx}\,x^2 + \frac{d}{dx}\,1\\[5pt]

Version vom 12:51, 10. Mär. 2009

We expand the quadratic expression as

\displaystyle \begin{align}

f(x) &= \bigl(x^2-1\bigr)^2\\[5pt] &= \bigl(x^2\bigr)^2 - 2\cdot x^2\cdot 1 + 1^2\\[5pt] &= x^4 - 2x^2 + 1\,\textrm{.} \end{align}

When the function is written in this form, it is easy to differentiate term by term,

\displaystyle \begin{align}

f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^4-2x^2+1\bigr)\\[5pt] &= \frac{d}{dx}\,x^4 - 2\frac{d}{dx}\,x^2 + \frac{d}{dx}\,1\\[5pt] &= 4\cdot x^{4-1} - 2\cdot 2x^{2-1} + 0\\[5pt] &= 4x^{3} - 4x\\[5pt] &= 4x(x^2-1)\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:2e“