Sideversjonen fra 19. jul. 2020 kl. 06:30

Løsningsforslag (pdf) (open source, meld fra om forbedringer eller feil her)

DEL 1

$f(x)=x^2+2x+e^x$

$f'(x)=2x+2+e^x$

$g(x)=x^2\cdot ln \, x$

$g'(x)=2x\cdot ln \, x + x^2 \cdot \frac{1}{x} = 2x\cdot ln\, x + x$

$h(x)=\frac{x-1}{e^{2x+1}}$

$h'(x)=\frac{1\cdot e^{2x+1}-(x-1)\cdot 2\cdot e^{2x+1}}{(e^{2x+1})^2} = \frac{1-(2x-2)}{e^{2x+1}} = \frac{-2x+1}{e^{2x+3}}$

@@ Linje 23: / Linje 23: @@
 $g(x)=x^2\cdot ln \, x$
-$g'(x)=2x\cdot ln \, x + x^2 \cdot frac{1}{x} = 2x\cdot ln\, x + x
+$g'(x)=2x\cdot ln \, x + x^2 \cdot \frac{1}{x} = 2x\cdot ln\, x + x$
 ===c)===
@@ Linje 29: / Linje 29: @@
 $h(x)=\frac{x-1}{e^{2x+1}}$
-$h'(x)=\frac{1\cdot e^{2x+1}-(x-1)\cdot 2\cdot e^{2x+1}}{(e^{2x+1})^2}$
+$h'(x)=\frac{1\cdot e^{2x+1}-(x-1)\cdot 2\cdot e^{2x+1}}{(e^{2x+1})^2} = \frac{1-(2x-2)}{e^{2x+1}} = \frac{-2x+1}{e^{2x+3}}$
+==Oppgave 2==