# Chain rule for function of several variables

Say we have a function $$f:X\times Y \to \mathbb{R}$$ where $$Y\subseteq \mathbb{R}^n$$. I want to calculate $$\frac{\partial}{\partial t}f(x,y_0+tu_i)$$ for a fixed $$y_0 \in Y$$ and $$u_i\in \mathbb{R}^n$$ is the i'th standard basis vector for a fixed $$i$$ and $$t\in \mathbb{R}$$. Now, just using the chain rule I think we get $$\frac{\partial}{\partial t}f(x,y_0+tu_i)=\frac{\partial f}{\partial y}(x,y_0+tu_i)u_i$$ But we haven't assumed that $$f$$ is differentiable. We only assume the existence of the partial derivative of $$f$$ w.r.t the $$i$$'th coordinate of $$y$$, so is the above even well-defined?

The important point is not to use the chain rule (since you don't know $$f$$ is differentiable), but just the definitions. If you're calculating this partial derivative with respect to $$t$$, you have fixed $$x=x_0$$ as well. Note that if we set $$t=s+t_0$$, we have $$\frac d{dt}\Big|_{t=t_0} f(x_0,y_0+tu_i) = \frac d{ds}\Big|_{s=0} f(x_0,(y_0+t_0u_i)+su_i) = \frac{\partial f}{\partial y_i}(x_0,y_0+t_0u_i).$$ (The final equality is just the definition of the partial derivative.)
If you don't know that $$f(x_0,\cdot)$$ is a differentiable function (of $$y$$), then what you've written doesn't actually make sense.
• I think I understand, but just to be sure: By definition, the partial derivative is $$\lim_{h\to 0} \frac{f(x_0,y_0+(t_0+h)u_i)-f(x_0,y_0+t_0u_i)}{h}$$ $$=\lim_{h\to 0} \frac{f(x_0,y_0+t_0u_i+hu_i)-f(x_0,y_0+t_0u_i)}{h}$$ $$=\frac{\partial f}{\partial y_i}(x_0,y_0+t_0u_i)$$ Because the second expression is the definition of the partial derivative w.r.t $y_i$? Jan 22, 2022 at 19:44