Kedjeregeln

Sats. (Kedjeregeln i <math>\mathbf{R}^2</math>)

• <math>f:\mathbf{R}^2 \longrightarrow \mathbf{R}</math> differentierbar
• <math>g:\mathbf{R} \longrightarrow \mathbf{R}^2, t \longmapsto (x,y)</math> deriverbar i <math>]\alpha,\beta[</math> med <math>V_g \subset D_f</math>

Då är <math>h(t) = f \circ g = f(x(t),y(t))</math> deriverbar i <math>]\alpha,\beta[</math> med <math>h'(t) = f'_x(x,y)x' + f'_y(x,y)y'</math>.

BEVIS.

<math>\frac{h(t+\Delta t) - h(t)}{\Delta t} = \frac{1}{\Delta t}\left( f(x(t+\Delta t),y(t+\Delta t)) - f(x(t),y(t)) \right) = [ f diff. ] = \frac{f'_x(x,y)(x(t+\Delta t) - x(t)) + f'_y(x,y)(y(t+\Delta t) - y(t)) + \sqrt{(x(t+\Delta t) - x(t))^2 + (y(t+\Delta t) - y(t))^}\rho(x(t+\Delta t) - x(t),y(t+\Delta t) - y(t)){\Delta t} = f'_x(x,y)x' + f'_y(x,y)y' + \sqrt{\left(\frac{x(t+\Delta t) - x(t)}{\Delta t}\right)^2 + \left(\frac{y(t+\Delta t) - y(t)}{\Delta t}\right)^2}}\rho(x(t+\Delta t) - x(t), y(t+\Delta t) - y(t)) \longrightarrow f'_x(x,y)x' + f'_y(x,y)y'</math> då <math>\Delta t \longrightarrow 0</math>, ty <math>\rho(h,k) \longrightarrow 0</math>. HELVETE!!!!!!!