matematikk.net

Kan noen løse de to derivasjons oppgavene her? sjekke om jeg fikk rett

For den første bruker du kvotientregelen:

f'(x)=[(2x+1)'*(x^2+2x+1)-(2x+1)*(x^2+2x+1)']/(x^2+2x+1)^2

=[2(x^2+2x+1)-(2x+1)(2x+2)]/(x^2+2x+1)^2

=(-2x^2-2x)/(x^2+2x+1)^2=-2x(x+1)/(x+1)^4=-2x/(x+1)^3


Den andre (husk kv.rot(t)=t^(1/2):

f'(t)=2*1/2*t^(-1/2)+2*(-1/2)*t^(-3/2)

=1/kv.rot(t)-1/(kv.rot(t^3))=1/kv.rot(t)-1/(t*kv.rot(t))

[tex]* f^{\prime}(x) \;=[/tex]

[tex]= \; \frac{(2x \:+\: 1)^{\prime}(x \:+\: 1)^2 \:-\: (2x \:+\: 1)[(x \:+\: 1)^2]^{\prime}}{(x \:+\: 1)^4}[/tex]

[tex]=\; \frac{2(x \:+\: 1)^2 \:-\: (2x \:+\: 1)[2(x \:+\: 1)]}{(x \:+\: 1)^4}[/tex]

[tex]=\; \frac{2(x \:+\: 1) \:-\: 2(2x \:+\: 1)}{(x \:+\: 1)^3}[/tex]

[tex]=\; \frac{2x \:+\: 2 \:-\: 4x \:-\: 2}{(x \:+\: 1)^3}[/tex]

[tex]=\; - \: \frac{2x}{(x \:+\: 1)^3}\, .[/tex]

[tex]* f^{\prime}(t) \;=\; \Big( 2t^{1/2} \:+\: 2t^{-1/2} \Big)^{\prime} \;=\; t^{-1/2} \:-\: t^{-3/2} \;=\; \frac{1}{\sqrt{t}} \:-\: \frac{1}{t\sqrt{t}}.[/tex]