R1 2009 vår LØSNING

Del 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)} </tex>

a)

<tex>\bigtriangleup ABC \sim \bigtriangleup ADC </tex>

Fordi vinkel A er den samme i begge trekanter og vinkel C (i ABC) er lik vinkel D (i ADC).

<tex>\bigtriangleup ABC \sim \bigtriangleup BCD </tex>

Fordi vinkel B er den samme i begge trekanter og vinkel C (i ABC) er lik vinkel D (i BCD).

b)

<tex> \frac{AC}{AB} = \frac{AD}{AC} \\ (AC)^2 = AD \cdot AB </tex>

<tex> \frac{BC}{AB} = \frac{BD}{BC} \\ (BC)^2 = BD \cdot AB</tex>

c)

<tex> (AC)^2 = AD \cdot AB \\ (BC)^2 = BD \cdot AB \\ </tex>