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A4-format til udskrift. - Aarhus Universitet

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38 I. DIFFERENTIATION<br />

Løsning<br />

Funktionerne g1 = x 2 − y 2 , g2 = xy med<br />

giver Jacobimatrix<br />

Afkryds den rigtige:<br />

g1x = 2x, g1y = −2y, g2x = y, g2y = x<br />

<br />

2x −2y<br />

=<br />

.<br />

y x<br />

g1x g1y<br />

g2x g2y<br />

(a) (b) (c)<br />

<br />

4.23. Implicit given funktion ☞ [S] 11.5 The chain rule<br />

Implicit funktion<br />

Ligningen<br />

F(x,y) = 0, Fy(a,b) = 0<br />

definerer en løsningsfunktion y(x) med F(x,y(x)) = 0 for x <strong>til</strong>pas nær a.<br />

Kædereglen giver<br />

Fxx ′ + Fyy ′ = 0<br />

og deraf<br />

6 y ′ (x) = − Fx<br />

4.24. Kurve er graf ☞ [S] 11.5 The chain rule<br />

Figur<br />

y<br />

1<br />

Fy<br />

F(x,y) = 0<br />

4.25. Inddirekte beregning ☞ [S] 11.5 The chain rule<br />

Eksempel 8<br />

F = x 3 + y 3 − 6xy = 0<br />

Fx = 3x 2 − 6y, Fy = 3y 2 − 6x<br />

når Fy = 3y 2 − 6x = 0.<br />

dy<br />

dx<br />

= −Fx<br />

Fy<br />

= − 3x2 − 6y<br />

3y2 − 6x<br />

= 2y − x2<br />

y2 − 2x<br />

x

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