What is $\frac{\partial }{\partial x_1}$ equal to if we have $y_1=\frac{x^2_1}{x_2},\:y_2=log\left(x_1\right)$? We got:
$$\frac{\partial }{\partial x_1}=\frac{\partial y_1\:}{\partial \:x_1}\frac{\partial \:}{\partial \:y_1}+\frac{\partial \:y_2}{\partial \:x_1}\frac{\partial \:}{\partial \:y_2}$$
But isn't this wrong? Because it would mean that:
$$\frac{\partial }{\partial x_1}=\frac{\partial \:}{\partial \:x_1}+\frac{\partial \:}{\partial \:x_1}=2\frac{\partial \:}{\partial \:x_1}$$
 A: $\frac{\partial }{\partial x_1}$ is an operator that works the same way as the derivative in one variable but for functions of several variables. For example if $f(x_1,x_2)=x_1^2+x_1x_2+x_2^3$ then $\frac{\partial }{\partial x_1}f(x_1,x_2)=2x_1+x_2$ ,  $x_2$ is considered a constant when you compute the derivative in order to $x_1$. Note now that if $f(x_1,x_2)=1$ then $\frac{\partial }{\partial x_1}f(x_1,x_2)=0$ because it does not depend on $x_1$.
Writing $\frac{\partial }{\partial x_1}f(x_1,x_2)$ where $f(x_1,x_2)=1$ differently with the help of $\frac{\partial }{\partial x_1}y_1(x_1,x_2)$ and $\frac{\partial }{\partial x_1}y_2(x_1,x_2)$:
$\frac{\partial }{\partial x_1}y_1(x_1,x_2)=2\frac{x_1}{x_2}  \text{ and } \frac{\partial }{\partial x_1}y_2(x_1,x_2)=\frac{1}{x_1}$
$$\text{Therefore } 0=\frac{\partial }{\partial x_1}f(x_1,x_2)=0\times\frac{\partial }{\partial x_1}y_1(x_1,x_2)+0\times\frac{\partial }{\partial x_1}y_2(x_1,x_2)$$
Now imagine that $x_1=x_1(t) $ and $x_2=x_2(t)$ then
$\frac{d }{dt}y_1(x_1,x_2)=\frac{\partial }{\partial x_1}y_1(x_1,x_2)\frac{d}{dt}x_1(t)+\frac{\partial }{\partial x_2}y_1(x_1,x_2)\frac{d}{dt}x_2(t)$
