Let $g=g(x,y)$ be a differentiable real-valued function and let $a,b\colon\mathbb R\to \mathbb R$ be differentiable. Prove that $$\left.\frac{d}{dt}g(a(t),b(t))\right|_{t=0}=\frac{\partial g}{\partial x}(a(0),b(0))a'(0)+\frac{\partial g}{\partial y}(a(0),b(0))b'(0).$$
I tried as follow :
$$\frac{g(a(t),b(t))-g(a(0),b(0))}{t}=\frac{g(a(t),b(t))-g(a(0),b(t))+g(a(0),b(t))-g(a(0),b(0))}{t}$$
I know that $$\frac{g(a(0),b(t))-g(a(0),b(0))}{t}$$ $$=\frac{g(a(0),b(t))-g(a(0),b(0))}{b(t)-b(0)}\cdot \frac{b(t)-b(0)}{t}\to \frac{\partial g}{\partial y}(a(0),b(0))b'(0)$$ when $t\to 0$, but how can I prove that $$\frac{g(a(t),b(t))-g(a(0),b(t))}{t}$$ $$=\frac{g(a(t),b(t))-g(a(0),b(0))}{a(t)-a(0)}\cdot \frac{a(t)-a(0)}{t}\to \frac{\partial g}{\partial x}(a(0),b(0))a'(0) \ \ ?$$
My problem is to prove that $$\lim_{t\to 0}\frac{g(a(t),b(t))-g(a(0),b(t))}{a(t)-a(0)}=\frac{\partial g}{\partial x}(a(0),b(0)),$$ because the seconde variable in $g(a(t),b(t))$ moves when $t\to 0$. Any idea ?
I'm not sure that you can conclude that way. Since $g$ is differentiable at $(a(0),b(0))$, then, $$g(x,y)=g(a(0),b(0))+\nabla g(a(0),b(0))\cdot ((x,y)-(a(0),b(0))+o((x,y)-(a(0),b(0))).$$ Therefore, $$\frac{g(a(t),b(t))-g(a(0),b(0))}{t}=\nabla g(a(0),b(0))\cdot \left(\frac{a(t)-a(0)}{t},\frac{b(t)-b(0)}{t}\right)+o(1).$$ Taking $t\to 0$, the claim follow.