From Förberedande kurs i matematik 2

Revision as of 08:32, 4 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_1c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (13:51, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
(3 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	The expression is a quotient of two polynomials, <math>x^2+1</math> and <math>x+1</math>, and we therefore use the quotient rule for differentiation,
-	<center> [[Bild:1_2_1c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	\Bigl(\frac{x^2+1}{x+1}\Bigr)'
		+	&= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt]
		+	&= \frac{2x\cdot (x+1) - (x^2+1)\cdot 1}{(x+1)^2}\\[5pt]
		+	&= \frac{2x^2+2x-x^2-1}{(x+1)^2}\\[5pt]
		+	&= \frac{x^2+2x-1}{(x+1)^2}\,\textrm{.}
		+	\end{align}</math>}}
		+
		+
		+	Note: It is possible to rewrite the numerator by completing the square,
		+
		+	{{Displayed math||<math>x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2</math>}}
		+
		+	and then the answer can be written as
		+
		+	{{Displayed math||<math>\frac{x^2+2x-1}{(x+1)^2} = \frac{(x+1)^2-2}{(x+1)^2} = 1-\frac{2}{(x+1)^2}\,\textrm{.}</math>}}

---

Current revision

The expression is a quotient of two polynomials, \displaystyle x^2+1 and \displaystyle x+1, and we therefore use the quotient rule for differentiation,

\displaystyle \begin{align} \Bigl(\frac{x^2+1}{x+1}\Bigr)' &= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt] &= \frac{2x\cdot (x+1) - (x^2+1)\cdot 1}{(x+1)^2}\\[5pt] &= \frac{2x^2+2x-x^2-1}{(x+1)^2}\\[5pt] &= \frac{x^2+2x-1}{(x+1)^2}\,\textrm{.} \end{align}

Note: It is possible to rewrite the numerator by completing the square,

\displaystyle x^2+2x-1 = (x+1)^{2} - 1^2 - 1 = (x+1)^2 - 2

and then the answer can be written as

\displaystyle \frac{x^2+2x-1}{(x+1)^2} = \frac{(x+1)^2-2}{(x+1)^2} = 1-\frac{2}{(x+1)^2}\,\textrm{.}