From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	We differentiate term by term,
-	<center> [[Image:1_1_2c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(e^x-\ln x\bigr)\\[5pt]
		+	&= \frac{d}{dx}\,e^{x} - \frac{d}{dx}\,\ln x\\[5pt]
		+	&= e^{x}-\frac{1}{x}\,\textrm{.}
		+	\end{align}</math>}}
		+
		+
		+	Note: Because <math>\ln x</math> is not defined for <math>x\le 0</math> we assume implicitly that <math>x > 0</math>.

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Current revision

We differentiate term by term,

\displaystyle \begin{align} f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(e^x-\ln x\bigr)\\[5pt] &= \frac{d}{dx}\,e^{x} - \frac{d}{dx}\,\ln x\\[5pt] &= e^{x}-\frac{1}{x}\,\textrm{.} \end{align}

Note: Because \displaystyle \ln x is not defined for \displaystyle x\le 0 we assume implicitly that \displaystyle x > 0.