Let $f:\mathbb{R}^2\to\mathbb{R}$ be a differential function at the origin, and $\gamma_1,\gamma_2: (-1,1)\to \mathbb{R}^2$ continously differentiable curves at the origin, s.t. $\gamma_1(0)=\gamma_2(0)=(0,0)$, and $\forall t\in(-1,1) :\gamma_2(t)=\gamma_1(t)+(t^2,t^3)$.
Show $\frac {d (f(\gamma_1(t))} {dt}\bigg\rvert_{t=0} = \frac {d (f(\gamma_2(t))} {dt} \bigg\rvert_{t=0}$.
I tried applying the chain rule the following way:
1)$D_{f\circ\gamma_1(t)}(0)=D_f(0)D_{\gamma_1}(0)$.
2) $D_{f\circ\gamma_2(t)}(0)=D_f(0)D_{\gamma_2}(0)=D_f(0)D_{\gamma_1+(t^2,t^3)}(0)=^*D_f(0)(D_{\gamma_1}+(2t,3t^2)(0)=D_f(0)D_{\gamma_1}(0)$
(*)- From linearity of the derivative operation
Is this use of the chain rule correct? And if so, where did I use continuity of the derivatives of $\gamma$?
It's correct. A simpler argument would be to note that $D_{\gamma_1}(0)=D_{\gamma_2}(0)$. This, together with a pointer to the chain rule, proves the claim without any calculations.