Intuitive Proof of the Chain Rule in 1 Variable Is there a simple and intuitive way to prove the chain rule, that is, if $y$ is a function of $u$ and $u$ is a function of $x$, then why is $\frac{dy}{dx}$ = $\frac{dy}{du}$ $\cdot$ $\frac{du}{dx}$ ? This could just be an intuitive argument. 
PS: The only proofs I found were based off of confusing definitions.
 A: 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".
A: 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.
A: 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., $(g\circ f)'(a) = g'(f(a))\cdot f'(a).$
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.
A: The best intuition, in my opinion, comes from the notion of a differential. To each scalar variable $v$, there is a corresponding differential $\mathrm{d}v$.
Among the things you can do with differentials are:


*

*Add them

*Multiply a differential by a scalar


If it turns out two differentials are related by an equation
$$\mathrm{d}v = w \, \mathrm{d}u $$
then $w$ is* determined this equation, and
it makes sense to define the ratio
$$ \frac{\mathrm{d}v}{\mathrm{d}u} = w$$
Of course, if $f$ is differentiable, then we have related differentiables
$$\mathrm{d}(f(u)) = f'(u) \, \mathrm{d}u $$ 
If all of the ratios involved are defined, we can compute
$$ \frac{\mathrm{d}y}{\mathrm{d}x} \, \mathrm{d}x
= \mathrm{d}y = \frac{\mathrm{d}y}{\mathrm{d}u} \mathrm{d}u
= \frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x} \mathrm{d}x$$
and conclude
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x} $$
*: There are caveats involved; e.g. this equation doesn't tell us anything about $w$ in a region where $u$ is locally constant (and so $\mathrm{d}u = 0$)

This sort of calculation only works well in one dimension; e.g. where all of the variables involved are related. When you have multiple independent variables, differentials still make sense, but ratios usually don't. 
For example, on the plane, $\mathrm{d}x$ and $\mathrm{d}y$ are both well-defined, but neither is a scalar multiple of the other; one can't make sense out of "dividing": by $\mathrm{d}x$, except in special cases.
A: $\frac {f(g(x))}{d(g(x))}*\frac{d(g(x))}{dx}=$
$\lim_{H_h = g(x+h)- g(x)\rightarrow 0} \frac {f(g(x) + H_h) -f(g(x)) }{H_h}*\lim_{h= x+h - x = h\rightarrow 0} \frac{g(x+h) - g(x)}h$
$= \lim_{h\rightarrow 0}\frac {f(g(x) + H_h) -f(g(x)) }{H_h}*\lim_{h\rightarrow 0} \frac{g(x+h) - g(x)}h$
$= \lim_{h\rightarrow 0}\frac {f(g(x) + g(x+h) - g(x)) -f(g(x)) }{H_h}* \frac{g(x+h) - g(x)}h$
$= \lim_{h\rightarrow 0}\frac {f(g(x+h)) -f(g(x)) }{H_h}* \frac{H_h}h$
$= \lim_{h\rightarrow 0}\frac {f(g(x+h)) -f(g(x)) }h$
$= \frac {d(f(g(x))}{dx}$
(See comments.  If $H_h = g (x+h)-g(x)=0$ for all $0 < h < \delta $ for some $\delta $ then those limits don't really work.)
(But then $g'(x) = \lim\frac {g (x+h)-g (x)}h=\lim \frac 0h =0$ and $g (x)=g (x+h) $ so $\frac {d(f (g))}{dx}=\lim \frac {f (g (x+h))-f (g (x))}h=\lim\frac {f (g(x))-f (g(x))}h=\lim \frac 0h=0 = \frac {dfg (x)}{dg (x)}*0=\frac {dfg (x)}{dg (x)}*\frac {dg (x)}{dx} $ so that's trivial.)
===
This is highly abusive, but for small values of $h \ne 0 $:
$f'(x)\approx \frac {f (x+h)-f (x)}h $ so
$hf'(x)\approx f (x+h) - f (x) $
$f (x)\approx f (x+h)-hf'(x) $ and
$f (x+h)\approx f (x)+hf'(x)$
and thus all hold true even for $h=0$ (but they are rough)
And so for small values of $h $ and small values of $k=hg'(x) $ 
$h(f\circ g)'(x)\approx$
$f (g (x+h))-f (g (x))\approx $
$f (g (x)+hg'(x))-f (g (x))\approx $ (for small $k=hg'(x) $
$kf'(g (x))\approx $
$hg'(x)f '(g (x))$
So $(f\circ g)'(x)\approx g'(x)f' (g (x)) $.
Lots more rigor is required to make those valid limit expressions ($h $ can't equal $0$ but $g'(x) $ might, etc.)  but that is intuitive.
A: I think this proof is intuitive.
$$g(x) = G(f(x))$$
let $y = G(t)$, $t = f(x)$ and $y = g(x)$
$\Delta y, \Delta x ,\Delta t , \varepsilon$ are infinitesimals.
By increment theorem, 
$$\Delta y = G^\prime(t)\Delta t + \varepsilon\Delta t$$
Dividing by $\Delta x$
$${\Delta y \over \Delta x} = G^\prime(t){\Delta t\over \Delta x} + \varepsilon{\Delta t\over \Delta x}$$
Taking standard part, 
$$st\left({\Delta y \over \Delta x}\right) = G^\prime(t)\left({\Delta t\over \Delta x}\right) $$
$${\mathrm dy \over \mathrm d x}= G^\prime(t)\left({\mathrm dt\over \mathrm dx}\right) $$
$$\bbox[#F85, 5px, Border: 3px solid green] {g^\prime(x)= G^\prime(f(x))f^\prime (x)} $$
