Es sei gu(u,v)=∂∂ug(u,v)=ln(u⋅sin(u)−v){\displaystyle g_{u}(u,v)={\frac {\partial }{\partial _{u}}}g(u,v)=ln(u\cdot sin(u)-v)} und gv(u,v)=∂∂vg(u,v)=tan(−u+v3){\displaystyle g_{v}(u,v)={\frac {\partial }{\partial _{v}}}g(u,v)=tan(-u+v^{3})}. Man bestimme h(t)=ddtg(2t,t2+1){\displaystyle h(t)={\frac {d}{dt}}g(2t,t^{2}+1)}.
Formel: F′=guu′+gvv′{\displaystyle F'=g_{u}u'+g_{v}v'} für F(t)=g(u(t),v(t)){\displaystyle F(t)=g(u(t),v(t))}
u(t)=2t⇒u(t)′=2{\displaystyle u(t)=2t\Rightarrow u(t)'=2}
v(t)=t2+1⇒v(t)′=2t{\displaystyle v(t)=t^{2}+1\Rightarrow v(t)'=2t}
h(t)=gu(2t,t2+1)⋅u(t)′+gv(2t,t2+1)⋅v(t)′=ln(2t⋅sin(2t)−(t2+1))⋅2+tan(−2t+(t2+1)3)⋅2t{\displaystyle h(t)=g_{u}(2t,t^{2}+1)\cdot u(t)'+g_{v}(2t,t^{2}+1)\cdot v(t)'=ln(2t\cdot sin(2t)-(t^{2}+1))\cdot 2+tan(-2t+(t^{2}+1)^{3})\cdot 2t}
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