R1 2009 vår LØSNING: Forskjell mellom sideversjoner
Fra Matematikk.net
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== c) == | == c) == | ||
<tex> \frac{x-2}{x^2+2x}- \frac{x+2}{x^2-2x}-\frac{4x}{x^2-4} = \\ \frac{x-2}{x(x+2)}- \frac{x+2}{x(x-2)} - \frac{4x}{(x+2)(x-2)} = \\ \frac{(x-2)(x-2)-(x+2)(x+2)- 4x^2}{x(x+2)(x-2)} = \\ \frac{x^2-4x+4-(x^2+4x+4)- 4x^2}{x(x+2)(x-2)} = \\ \frac{- 4x}{(x+2)(x-2)}</tex> | <tex> \frac{x-2}{x^2+2x}- \frac{x+2}{x^2-2x}-\frac{4x}{x^2-4} = \\ \frac{x-2}{x(x+2)}- \frac{x+2}{x(x-2)} - \frac{4x}{(x+2)(x-2)} = \\ \frac{(x-2)(x-2)-(x+2)(x+2)- 4x^2}{x(x+2)(x-2)} = \\ \frac{x^2-4x+4-(x^2+4x+4)- 4x^2}{x(x+2)(x-2)} = \\ \frac{- 4x}{(x+2)(x-2)} \triangel</tex> | ||
Sideversjonen fra 27. mar. 2012 kl. 10:29
Del 1
Oppgave 1
a)
1)
<tex>f(x) = (x^2+1)^4 \\ f'(x)= 4(x^2+1)^3 \cdot 2x = 8x(x^2+1)^3</tex>
(kjerneregelen)
2)
<tex>g(x) = xe^{2x} \\ g'(x)= e^{2x}+xe^{2x} \cdot 2 = e^{2x}(1+2x)</tex>
(produktregelen)
b)
<tex>\lim_{x \to 2} \frac{x^2-2x}{x-2} =\lim_{x \to 2} \frac{x(x-2)}{x-2}=\lim_{x \to 2} x=2</tex>
c)
<tex> \frac{x-2}{x^2+2x}- \frac{x+2}{x^2-2x}-\frac{4x}{x^2-4} = \\ \frac{x-2}{x(x+2)}- \frac{x+2}{x(x-2)} - \frac{4x}{(x+2)(x-2)} = \\ \frac{(x-2)(x-2)-(x+2)(x+2)- 4x^2}{x(x+2)(x-2)} = \\ \frac{x^2-4x+4-(x^2+4x+4)- 4x^2}{x(x+2)(x-2)} = \\ \frac{- 4x}{(x+2)(x-2)} \triangel</tex>