Richtungsableitung/V nach C/Produktregel/Aufgabe/Lösung

Wir schreiben ${}f=f_{1}+{\mathrm {i} }f_{2}$ und ${}g=g_{1}+{\mathrm {i} }g_{2}$ . Unter Verwendung von Fakt und der reellen Version von Fakt ist

{}{\begin{aligned}D_{v}(fg)&=D_{v}{\left({\left(f_{1}+{\mathrm {i} }f_{2}\right)}{\left(g_{1}+{\mathrm {i} }g_{2}\right)}\right)}\\&=D_{v}{\left(f_{1}g_{1}-f_{2}g_{2}+{\mathrm {i} }{\left(f_{2}g_{1}+f_{1}g_{2}\right)}\right)}\\&=D_{v}{\left(f_{1}g_{1}\right)}-D_{v}{\left(f_{2}g_{2}\right)}+{\mathrm {i} }D_{v}{\left(f_{2}g_{1}\right)}+{\mathrm {i} }D_{v}{\left(f_{1}g_{2}\right)}\\&=f_{1}D_{v}g_{1}+g_{1}D_{v}f_{1}-f_{2}D_{v}g_{2}-g_{2}D_{v}f_{2}+{\mathrm {i} }f_{2}D_{v}g_{1}+{\mathrm {i} }g_{1}D_{v}f_{2}+{\mathrm {i} }f_{1}D_{v}g_{2}+{\mathrm {i} }g_{2}D_{v}f_{1}\\&={\left(f_{1}+{\mathrm {i} }f_{2}\right)}D_{v}{\left(g_{1}+{\mathrm {i} }g_{2}\right)}+{\left(g_{1}+{\mathrm {i} }g_{2}\right)}D_{v}{\left(f_{1}+{\mathrm {i} }f_{2}\right)}\\&=fD_{v}g+gD_{v}f.\end{aligned}}