Can anyone explain implicit differentiation in an intuitive way? I am in my final year of school. Understand the procedure and can do it well but do not understand why ? Can someone help ?
 A: Well, you use implicit differentiation where you are not able to write $y$ in terms of $x$. For example, let us say you have the equation:$$x^2 + y^2 = 25$$ One of the commonest examples you'll come across. This is the graph of a circle centred at origin and radius $\sqrt{25} = 5$. Notice that you can't write
$$\color{red}{y = \sqrt{25-x^2}}$$ since the graph only becomes a semicircle. Writing
$$\color{red}{y = \pm \sqrt{25-x^2}}$$ won't help either since now you'll get the derivative in $\pm$ form, but there is a single derivative at a point (given by the slope of the tangent to the circle at that point). This is where implicit differentiation comes in. We'll get to our original equation:
$$\color{green}{x^2 + y^2 = 25}$$ Consider the graph:

Every point on the circle will satisfy our equation. Thus, if we differentiate w.r.t $x$, then still any point on the circle will satisfy.
$${d\over dx}(x^2 + y^2) = {d\over dx}25$$
$${d\over dx}x^2 + {d\over dx}y^2 = 0$$
$$2x + {d\over dx}y^2 = 0$$ Here's the tricky part. $y$ is not just some plain old constant, it varies if you change $x$. Rewriting:
$$2x + {dy^2\over dy}\times {dy\over dx} = 0$$ This does not make any change to our equation as the $dy$ terms cancel out, or more formally, this abides by the chain rule.
We get:
$$2x + 2y{dy\over dx} = 0$$
$$x + y{dy\over dx} = 0$$
$${dy\over dx} = -{x\over y}$$
This is just plain old differential calculus, and we have solved it using the basic rules of differentiation.
Since you claim you understand the procedure, I won't explain these steps in a detailed way. 
Getting to your question: 
Now, we've arrived at the formula for $dy\over dx$ (actually this formula holds for any circle centred at the origin). In the above steps, I emphasised that any point on the circle will satisfy the equation. Understanding the equation, it reads the slope at $(x,y)$ is equal to $-{x\over y}$. So, you can plug in any $(x,y)$ (should be on the circle), to get the slope at that point. For example the slope at $(3,4)$ is $-0.75$ (verified by graphing).
Why this works:
Let us understand what we've done so far. We took our equation (which was not in the $y = f(x)$ form) and differentiated w.r.t $x$. Using the chain rule, we arrived at our answer. You might wonder why we haven't got the derivative as a function purely in terms of $x$ (this used to be my doubt too when I learnt this). The answer lies in the equation we started with: $x^2 + y^2 = 25$. Our original equation was not a function purely in $x$, so we are bound to get another one not purely in terms of $x$ as the derivative. Trying to understand it intuitively, if the derivative had been purely in terms of $x$, then you could plug in the value of $x$ to get the slope. But, notice that for every $x$ on the circle, there could be 2 corresponding $y$ values (and vice-versa). The slope at $(3,4)$ is $-0.75$, but at $(3,-4)$ is $0.75$. 
Okay, now we know why our derivative looks like this. Now, the equation reads: Slope = minus $x$ over $y$. So, for any $(x,y)$ on the circle, we can get the slope. Better yet, one can understand $${dx\over dy} = -{x\over y}$$ by rewriting in the Lagrange notation (this helped me a lot to understand the implicit differentiation). $$f'(x_y) = -{x \over y}$$ Where the derivative at $x$ is given by $f'(x)$, but we write it in the form of $x_y$ to distinguish the 2 corresponding $y$'s. For example $f'(3_4) = -0.75$ and $f'(3_{-4}) = 0.75$. In case of normal differentiation using Lagrange notation, suppose we get $f'(x) = g(x)$. Then we can get the derivative at $x = a$ by $g(a)$, by simply plugging in the value. The process is similar here, just, you plug the value of $x$ along with which $y$ you want.
I hope I explained it clearly.
A: There is a theorem called the Implicit Function Theorem which is usually taught in a course on real analysis. What this theorem states from the perspective of a calculus student is that if we have a two variable function of the form $F(x,y)=0$ that under some conditions, usually not stated in a calculus course, we can view $y$ as a function of $x$.
So, for example, if we have an expression like $$x-y^2=5$$ we can view that as $$F(x,y)=x-y^2-5$$ and since our original relation is $x-y^2=5$ we have that $$F(x,y)=0$$
The theorem tells us that we can assume $y$ can be expressed as a function of $x$.
Note that the theorem also tells us that we can view $y$ as the variable and $x$ as a function of $y$.
Viewing $y$ as a function of $x$, our multivariable function $F(x,y)$ becomes $$F(x,y(x))=x-y(x)^2-5$$ which is now a function in the single variable $x$. Therefore, to differentiate we use the chain rule to get $$F'(x,y(x))=1-2y\frac{dy}{dx}$$
But we also have that $F(x, y(x))=0$ and since the derivative of a constant is $0$ we get $$1-2y\frac{dy}{dx}=0$$ which is the form we see when we do implicit differentiation.
Since all of this is justified by the Implicit Function Theorem, we skip the definition of $F$ and since we understand that we are considering $y$ to be a function of $x$ we go directly from $$x-y^2=5$$ to $$1-2y\frac{dy}{dx}=0$$
