Can someone give me a deeper understanding of implicit differentiation? I'm doing calculus and I want to be an engineer so I would like to understand the essence of the logic of implicit differentials rather than just memorizing the algorithm. Yes, I could probably memorize it and get a 100% on a test, but it means nothing unless I understand it and can acquire a practical understanding of it. I would really appreciate if someone can enlighten me.

What I understand already:

I understand how the derivatives of normal functions are found the long way... i.e.

$ f'(x)=\frac{f(x+h)-f(x))}{(x+h)-x} $

I understand that the derivative of the above equation is found when we find the limit as h, representing the distance between the two points, approaches 0. 

I also understand that the chain rule, quotient rule, etc. are just algorithms that speed up the process of finding the derivative. I don't need a proof for that

An example problem

When we do implicit differential equations such as this one:

A ladder is 8.5 m long leaning against a wall, the bottom part of the ladder is 6.0 m from the wall and is sliding away from the wall at a rate of 2.5m/s. 

$x^2 + y^2 = h^2$ (Pythagorean theorem) (x is x value, y is y value, h is hypot)
We can find that y = 6.2m.

The derivative is 
$2x\frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t} = 2h\frac{\mathrm{d} h}{\mathrm{d} t}$

What I don't understand: (although I can do them by memorizing the algorithm)

Is the implicit differential... How do we relate all the terms to the change in time? I mean how do we know we can make that equation's derivative with respect to t? Would it be possible to make them all with respect to the change in x? If so please demonstrate, I think that would help a lot as my biggest lack of understanding is how to know what the bottom term should be for each derivative term. 

I really appreciate any help, thanks.
Don't treat implicit differentiation as an idea that is distinct from "regular" differentiation.  Just as the chain rule is involved in every derivative, and is not a separate rule that governs only certain situations, e.g.
$$y=x^2=(x)^2$$
$$
\begin{align}
\frac{d}{dx}(y)&amp;=\frac{d}{dx}(x)^2\\
&amp;=2(x)^1\frac{d}{dx}(x)\\
&amp;=2x\cdot 1\\
&amp;=2x
\end{align}
$$
implicit differentiation is something that you're doing all the time, you just don't see it as such.  For example, when you differentiate $y=3x^2+2$, view it as differentiating both sides with respect to the symbol $x$:
$$
\begin{align}
\frac{d}{dx}(y)&amp;=\frac{d}{dx}(3x^2+2)\\
&amp;=\frac{d}{dx}(3x^2)+\frac{d}{dx}(2)\\
&amp;=3\frac{d}{dx}(x^2)+0\\
&amp;=3\cdot 2(x)^1\frac{d}{dx}(x)\quad\text{ (via chain rule)}\\
&amp;=6\cdot x\cdot 1\\
&amp;=6x
\end{align}
$$
I think that there are two things that aid in understanding, both of which treat differentiation in a symbolic manner.  


The first is to realize that nothing changes if the equation is rearranged, for example if we change it to $y-3x^2=2$, the same process works.  Differentiating both sides with respect to the symbol $x$ gives:
$$
\begin{align}
\frac{d}{dx}(y-3x^2)&amp;=\frac{d}{dx}(2)\\
\frac{d}{dx}(y)-\frac{d}{dx}(3x^2)&amp;=\frac{d}{dx}(2)\\
\frac{d}{dx}(y)-6x\frac{d}{dx}(x)&amp;=0\\
\frac{dy}{dx}&amp;=6x
\end{align}
$$
The second is to realize (or treat) $x$ as simply being an arbitrary symbol. That is, think of $\dfrac{d}{dx}$ as $\dfrac{d}{d\square}$.  The operations above work just the same.  We would end up with:
$$
\frac{d}{d\square}(y)-6x\frac{d}{d\square}(x)=0
$$
When, if $\square =x$ we get (as above)
$$
\frac{dy}{dx}=6x
$$
while if $\square=y$ we get
$$
\begin{align}
\frac{d}{dy}(y)-6x\frac{d}{dy}(x)&amp;=0\\
1-6x\frac{dx}{dy}&amp;=0\\
\frac{dx}{dy}&amp;=\frac{1}{6x}
\end{align}
$$
Finally, if $\square=t$ then no "cancellation" occurs and we get
$$
\frac{d}{dt}(y)-6x\frac{d}{dt}(x)=0
$$


The only other thing to remember is that the chain rule needs to be applied to all symbols as in:
$$
\frac{d}{d\square}(y)^2=2(y)^1\frac{d}{d\square}(y)
$$
Given: dx/dt = 2.5 m/s ; h = 8.5 m ; x = 6.0 m

Because the ladder is assumed to be constrained to the surfaces - the Pythagorean relationship holds at any given instance in time or space. 

Assume ladder doesn't stretch and is rigid so its length (hypotenuse) doesn't change with time. dh/dt = 0 

Using d/dt:

Step (1) x*dx/dt = -y*dy/dt

Step (2) find y when x = 6.0 m and h = 8.5 m;

Step (3) plug x, y and dx/dt into (1) and solve dy/dt = -2.4 m/s

Now to answer your question. Assume dh/dx = 0 since it's not stretching

d/dx of our geometric constraint is: 2*x + 2*y*dy/dx = 0 

Using d/dx: 

Step (1) x  = -y*dy/dx ;

Step (2) again find y when x = 6.0 m and h = 8.5 m and plug x, y into above equation to get dy/dx = -6/6.2

Step (3) dy/dt = (dy/dx)*(dx/dt) ;// relationship defining dy/dt (chain rule again) 

Step (4) plug in dy/dx into (3) and plug in dx/dt(given)

Step (5) dy/dt = -6/6.2 * 2.5 = -2.4 m/s

Since we were given a rate which is defined as m/s it only makes sense here to take derivative w/ respective to time b/c as you see it involves less steps but you can solve it either way.