

Uses of implicit differentiation - pencil

Hello HN,<p>I have now gained an intuition of solving a simple equation involving implicit differentiation.example x^2y = 5 where y is considered to be a function of x even though it's not quite obvious.(it's not explicitly stated like y = f(x).)in this case we proceed taking the derivative using the power and the chain rule.my question is what is this implicit differentiation all about?i tried doing a google search on this but couldn't find a convincing article.wikipedia too starts off with the definition of it which i couldn't understand.neither of the two talks about the practicle uses of it.
i'll be really happy if someone could explain this concept in detail.so far i've been blindly solving problems involving this technic<p>Thank You
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SandB0x
In your example, y _is_ a function of x. x is also a function of y. If you
pick a value for x, there is a unique y that satisfies the equality. If you
pick a value for y, there a unique x that satisfies the equality. Let's find y
as a function of x

    
    
        x^2y = 5 => y = 5x^-2 <------ this is y(x)
    

That's all it means. This a _function_ , because as mentioned, x uniquely
determines y.

Now let's look at an example where y is _not_ a function of x - the standard
example of a circle of radius r:

    
    
        x^2 + y^2 = r^2
    

solving for y gives

    
    
        y = +- sqrt(r^2 - x^2).
    

This is NOT a function, because for a value of x, we don't know which value of
y to choose - it could be the positive case or the negative case.
Geometrically, this means a line (like y = c) can intersect the circle in two
places, so trying to find _the_ single intersection point is not well defined.

But the _gradient_ y' IS a function, a function of both x and y and we can
find it by implicit differentiation (with respect to x):

    
    
        2x + 2yy' = 0 => y' = -x/y  <---- this is y'(x,y)
    

Geometrically this means that _given_ a point (x,y) on a circle, the gradient
_is_ a well defined quantity, even though y(x) is not.

\------------

Also: I had a look at your submission history and I really don't think that HN
is the right place for most of your questions. I'm just procrastinating by
answering this. You should probably be using math.stackexchange.com or a
similar site.

~~~
SandB0x
Ugh, I shouldn't try and post at stupidly late hours. In case you're still
reading, in the first paragraph x is NOT also a function of y, as x = +-
sqrt(5/y), and we have the choice of signs to deal with. The rest is ok.

------
btilly
This is a perfect example of how math is very simple. It just isn't always
simple in a way our brains are wired to naturally follow. With the result that
when people "get" a concept, they beat themselves up for not getting something
so simple.

Implicit differentiation is simple in that way.

Let's take your example. xxy = 5. This implicitly defines some function y. xxy
is an expression. And 5 is another expression. And the two are equal for all
x. Now note that you can differentiate any expression you want to. For
instance the derivative of xxy is 2xy + xxy'. And the derivative of 5 is 0.

The entire idea of implicit differentiation is this. If two expressions are
the same for all x, then they have the same tangent lines, and therefore the
same derivatives. Thus you don't need to solve for y first. You can just
differentiate and solve for y' instead. This is convenient when y is difficult
to solve for.

In this case you find that 2xy + xxy' = 0 so xxy' = -2xy and so y' = -2y/x.

That's all there is to it.

Now two notes of interest. The first is that regular differentiation is really
just a special case of implicit differentiation. We get y=f(x) and
differentiate both sides to get y'=f'(x). The second is that equations like y'
= -2y/x are the start of a very important subject called differential
equations.

------
ggchappell
Your question is a little vague, but I'll try.

What it is: Differentiate both sides of an equation. That's it.

Why it works: When two things are equal, if we transform them in identical
ways (being careful about what "identical" means), then the results are equal.
So if two differentiable functions are equal, then their derivatives must be
equal.

What it's good for: In spite of all the carefully selected examples we're
given in calculus class, the fact is that solving equations is often very
difficult, and usually _impossible_. So if someone writes some "f(x,y) = k",
and asks you to solve for y in terms of x, well, in general, you can't do it.

Now, suppose you want to find dy/dx in the above situation. The "usual" way is
to solve for y and then differentiate. But what if you cannot do that? If you
can find the derivative, w.r.t. x, of f(x,y), then you can apply implicit
differentiation, to get something like g(x,y)*dy/dx + h(x,y) = 0. Then finding
dy/dx is easy.

So: What it is good for is finding a derivative while avoiding the pain (or
the impossibility) of solving an equation.

Of course, that leads to the question of what derivatives are good for, but
I'm hoping you have some inkling of the answer to that one. If not, then back
up, and learn more about derivatives.

