From Förberedande kurs i matematik 2

Revision as of 14:24, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 1.2:2b moved to Solution 1.2:2b: Robot: moved page) ← Previous diff		Current revision (08:02, 15 October 2008) (edit) (undo) Tek (Talk | contribs) m
(One intermediate revision not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	The whole expression consists of two parts: the outer part, "''e'' raised to something",
-	<center> [[Image:1_2_2b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,</math>}}
		+
		+	where "something" is the inner part <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x</math>.
		+
		+	The derivative is calculated according to the chain rule by differentiating <math>e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}</math> with respect to <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}</math> and then multiplying by the inner derivative <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)'</math>, i.e.
		+
		+	{{Displayed math||<math>\frac{d}{dx}\,e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}} = e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,x^2+x\,} \bigr)'\,\textrm{.}</math>}}
		+
		+	The inner part is an ordinary polynomial which we differentiate directly,
		+
		+	{{Displayed math||<math>\frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}</math>}}

---

Current revision

The whole expression consists of two parts: the outer part, "e raised to something",

\displaystyle e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,

where "something" is the inner part \displaystyle \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x.

The derivative is calculated according to the chain rule by differentiating \displaystyle e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}} with respect to \displaystyle \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} and then multiplying by the inner derivative \displaystyle \bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)', i.e.

\displaystyle \frac{d}{dx}\,e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}} = e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,x^2+x\,} \bigr)'\,\textrm{.}

The inner part is an ordinary polynomial which we differentiate directly,

\displaystyle \frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}