3 corrected an error

For $t\ne 0$ one has $${f(x,t)-f(x,0) \over t}- \partial_2 f(0,0)= \int_0^1 (\partial_2 f(x,\tau \thinspace \thinspace t) - \partial_2 f(0,0))\thinspace d\tau ,$$ and here the right side is $<\epsilon$ when $(x,t)$ is in a suitable neighbourhood of $(0,0)$.

For an $f:X\times {\bf R}^n\to {\bf R}^m$ it is enough to consider the $i$-th coordinate function $f_i:X\times {\bf R}^n\to {\bf R}$, again denoted by $f$, and for the latter consider the auxiliary function $\phi(t):=f(x,t \thinspace v)$ on the interval $[0,1]$. One gets $$f(x,v)-f(x,0) =\int_0^1 \phi'(t)\thinspace dt = \int_0^1 df_2(x,t\thinspace \nabla f_2(x,t\thinspace v)\cdot v\thinspace v \thinspace dt,$$ whence $$\lim_{(x,v)\to(x_0,0)} {f(x,v)-f(x,0) \over \|v\|} f(x,v)-f(x,0) = df_2(x_0,0)\cdot {v\over \|v\|}.$$ nabla f_2(x_0,0)\cdot v + o(\|v\|) \qquad ((x,v)\to(x_0,0)).$$2 added 481 characters in body For t\ne 0 one has$${f(x,t)-f(x,0) \over t}- \partial_2 f(0,0)= \int_0^1 (\partial_2 f(x,\tau \thinspace \thinspace t) - \partial_2 f(0,0))\thinspace d\tau ,$$and here the right side is <\epsilon when (x,t) is in a suitable neighbourhood of (0,0). For an f:X\times {\bf R}^n\to {\bf R}^m it is enough to consider the i-th coordinate function f_i:X\times {\bf R}^n\to {\bf R}, again denoted by f, and for the latter consider the auxiliary function \phi(t):=f(x,t \thinspace v) on the interval [0,1]. One gets$$f(x,v)-f(x,0) =\int_0^1 \phi'(t)\thinspace dt = \int_0^1 df_2(x,t\thinspace v)\cdot v\thinspace dt,$$whence$$\lim_{(x,v)\to(x_0,0)} {f(x,v)-f(x,0) \over \|v\|} = df_2(x_0,0)\cdot {v\over \|v\|}.$$1 For t\ne 0 one has$${f(x,t)-f(x,0) \over t}- \partial_2 f(0,0)= \int_0^1 (\partial_2 f(x,\tau \thinspace \thinspace t) - \partial_2 f(0,0))\thinspace d\tau , and here the right side is $<\epsilon$ when $(x,t)$ is in a suitable neighbourhood of $(0,0)$.