1
$\begingroup$

The question is, why implicit differentiation applies to equations [ Because they're not functions ]
As we know, we can apply derivation to the functions, and I know that in implicit differentiation we consider $y$ as a function of $x$, but what that function is?
I had also asked another question which didn't get a satisfying answer for it, it is also related to this matter.

Why are we allowed to take the derivative of ellipse equation? [ Is there any geometric representation of it? ]

Please, please, please to infinity , don't give Wikipedia link of the theorems.I just wonder about those people who copy/paste the Wikipedia links, if Wikipedia was satisfying, I would have never asked my questions in here.

I will absolutely appreciate whoever gives enlightening piece of favor!

*As a side note, people around me, students, and TA's can't even understand the question.So, it has become really important for me to know the answer!

$\endgroup$
2
  • 1
    $\begingroup$ Why does something need to be a function in order for you to differentiate it? See math.stackexchange.com/questions/936206/… $\endgroup$
    – MrSlunk
    Commented Sep 25, 2014 at 0:01
  • $\begingroup$ The short answer is that differentiation is local, and the circle is a function of $x$ locally, except at the left and rightmost points. This is also why even implicitly, the derivative at the left and rightmost points blows up. $\endgroup$
    – davidlowryduda ♦
    Commented Sep 25, 2014 at 0:22

1 Answer 1

Reset to default
2
$\begingroup$

It's true that the question is not very clear. But look, there is something simple going on here. If we have an equation: $$\text{something} = \text{something else}$$

Then surely we also have the equation: $$\text{derivative}(\text{something}) = \text{derivative}(\text{something else})$$

So if you accept that we can differentiate one side of the equation, then you must accept that we can differentiate both sides and set them equal to one another.

Do you accept this? You say that you understand that we treat $y$ as a function of $x$, but you ask what that function is. This misses the point. Implicit differentiation is useful precisely when we don't know exactly what function we're dealing with. For example, suppose that I'm looking at this equation:

$$e^y \sin y = x^3 + x^x$$

Solving for $y$ would be a nightmare (if it were even possible). But we can still obtain information about the derivative of $y$, just by differentiating both sides using the chain rule (do this; it's a good exercise).

In some cases, we may be able to find the slope of a curve at certain points without ever knowing a parameterization of the curve.

All of this hinges on whether or not we can think of $y$ in the above equation as a function of $x$. So? Can we?

(Note: MrSlunk just posted a link to a nice answer that discusses this final question in detail.)

$\endgroup$
2
  • $\begingroup$ Indeed, I knew this all, but if you think of $y$ as a function of $x$ in the circle equation, then you get $y=\sqrt {25-x^2}$ which only gives the upper half of the plane( half circle ) but in the implicit differentiation it works for the part under $x$ axis.Still not satisfied, but thanks for your reply. $\endgroup$
    – FreeMind
    Commented Sep 25, 2014 at 1:41
  • 1
    $\begingroup$ @FreeMind See the link that MrSlunk provided. This is a complicated question in general, but for your specific example, the problem is that you chose some particular $y$. You could also have chosen $y= -\sqrt{25-x^2}$, which is a different function that also satisfies that equation. For example, if you want to determine the slope at a point in the upper half of the circle, you would choose your $y$, and if you wanted to determine the slope at a point in the lower half, you would use my $y$. But we don't need to pick a $y$ immediately, since differentiation is valid whichever $y$ we choose. $\endgroup$
    – Andrew Dudzik
    Commented Sep 25, 2014 at 1:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .