# Differentiability of implicit functions

I am studying calculus where I came across implicit differentiation. To that point we always tested differentiability first and then proceed to find its derivative. But my course module (book) made no allusion to differentiability of implicit functions.

For functions which cannot be algebraically expressed explicitly, how can we be sure that the function is indeed differentiable. Explain kindly in terms which I can understand.

• if a function cannot be expressed explicitly, you can still find out if it's differentiable at a particular point using the definition of derivative (i.e. proving that the limit exists) – Vasya May 22 '18 at 16:09

## 2 Answers

There is a theorem in Advanced Calculus texts called

(Implicit Function Theorem or Dini's Theorem )

which explains the conditions under which implicit differentiation is possible.

For example, the Advanced Calculus text by Witold Kosmala, has it on page $574.$ Your function $F(x,y)$ should be continuously differentiable on a neighborhood of the point $(a,b)$ with $F(a,b)=0$ and $F_y(a,b)\ne 0$.

Then you have $f'(x)= \frac {dy}{dx} = - \frac {F_x}{F_y}$ on some neighborhood $(a-r,a+r)$

One way is to assume that it is differentiable everywhere. Then attempt to differentiate. If it yields no contradiction or singularity anywhere, you have your result. Otherwise it is not differentiable in at least one point.