Derivation of function from $\mathbb{R}^6$ to $\mathbb{R}$

Question

Suppose $f:\mathbb{R}^6\rightarrow\mathbb{R}$. Why does the following equality holds:

$$\frac{d}{dt}|_{t=0}{f(x_1-t,x_2,x_3,x_4,x_5,x_6)}=-\frac{\partial f}{\partial x_1}$$

I think that is the chain rule, but I am confused if i have to use it. Can someone help me? Thanks a lot.

Answer 1

It is exactly a chain rule. But to be clear note that: $$\left.\frac{d}{dt}\right|_{t=0}{f(x_1-t,x_2,x_3,x_4,x_5,x_6)}=-\frac{\partial }{\partial x_1}f(x_1,x_2,x_3,x_4,x_5,x_6).$$

You take the derivative with respect to the 'first variable slot' regardless of what is sitting in it ($x_1-t$ in this case), then you do the chain rule and multiply by the derivative of what is sitting in the first variable slot ($-1$ in this case).

Answer 2

Hint : Use chain rule. set $u_1=x_1-t$ and $u_i=x_i\qquad i=2,...,6$ $$\frac{d}{dt}|_{t=0}{f(u_1,x_2,x_3,x_4,x_5,x_6)}=\sum_1^6\frac{\partial}{\partial u_i}|_{u_1=x_1}{f(u_1,u_2,u_3,u_4,u_5,u_6)}\frac{\partial u_i}{\partial t}=-\frac{\partial f}{\partial x_1}$$