Solution 1.1:2e

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
m (Robot: Automated text replacement (-[[Bild: +[[Image:))
Current revision (12:24, 14 October 2008) (edit) (undo)
m
 
(2 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
We expand the quadratic expression as
-
<center> [[Image:1_1_2e.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{{Displayed math||<math>\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}</math>}}
 +
 
 +
When the function is written in this form, it is easy to differentiate term by term,
 +
 
 +
{{Displayed math||<math>\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}</math>}}

Current revision

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}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-ImperialCollege/index.php/Solution_1.1:2e"