Intuitive Proof of the Chain Rule in 1 Variable

If $\frac{du}{dx}=k\neq 0$ at some $x$, then a first-order (that is: linear) approximation of $du$ close to $x$, gives $$ du=k\cdot dx \Rightarrow \frac{1}{du}=\frac{1}{k\cdot dx} $$ thus: $$ \frac{dy}{du}\cdot\frac{du}{dx}=\frac{dy}{k\cdot dx}\cdot k=\frac{dy}{dx} $$ Intuitively, you should be thinking of differentials as "small changes". So small, that even linear approximation would be good "enough".

If you want intuitive and simple:

$$\frac{dy}{dx}=\frac{dy}{\color{#4499de}{du}}\frac{\color{#4499de}{du}}{dx}$$

where the $du$'s cancel out.

If you want to be more rigorous, replace $dy,dx,du$ with $\Delta y,\Delta x,\Delta u$, which is the change with respect to $x$, and take the limit as $\Delta\to0$, which becomes the derivatives.

Some intuition: If $f(x) = m_1x + b_1, g(x) =m_2x + b_2,$ then $(g\circ f)(x) = m_2m_1x + (m_2b_1 + b_2).$ So in the case of linear functions, the slope of their composition is the product of the slopes. Now if $f,g$ are differentiable at $a, f(a)$ respectively, we can expect that, near $a,$ $g\circ f$ is very close to the composition of their tangent lines. Thus the slope of their composition at $a$ should be the product of the two slopes, i.e., $(g\circ f)'(a) = g'(f(a))\cdot f'(a).$

This answer is more intuition than an actual proof, but it may be helpful if you're learning the chain rule. The derivative of $y$ with respect to $x$ tells you how fast $y$ is changing as $x$ changes. If $x$ is changing 3 times as fast as $t$, and $y$ is changing 2 times as fast as $x$, then $y$ is changing 6 times as fast as $t$. This is not a proof but it gives you an idea of why it should be true. In the example above $dy/dx$ and $dx/dt$ are both constant, so $y(x)$ and $x(t)$ are linear functions, but for each value of $t$ and $x$, the graphs of $x(t)$ and $y(x)$ are "basically" lines.