I understand the method used in implicit differentiation, it's just an application of the chain rule. But why can you define $y$ as a function of $x$?

In this equation for example:
$x^2 + y^2 = 1$

$2x + 2yy' = 0 $

Why isn't it just this?:
$2x + 2y = 0$

Thank you in advance.

  • 2
    $\begingroup$ short answer: you differentiate with respect to $x$ not to a mixture of $x$ and $y$. $\endgroup$ – user251257 Feb 8 '17 at 23:25
  • $\begingroup$ another short answer: implicit differentiation involves you assuming that $y$ is a function of $x$ $\endgroup$ – Timothy Cho Feb 8 '17 at 23:26
  • $\begingroup$ @user251257 But why not assume y is constant when differentiating with respect to x? $\endgroup$ – SadSeven Feb 8 '17 at 23:35
  • $\begingroup$ @SadSeven: Why should it? If it is constant, then $\frac{d}{dx} x^2 + y^2 = 2x$. $\endgroup$ – user251257 Feb 8 '17 at 23:38
  • $\begingroup$ @SadSeven We treat $y$ as a constant in partial differentiation but not in ordinary differentiation. $\endgroup$ – projectilemotion Feb 8 '17 at 23:41

(1). $\{(x,y): x^2+y^2 =1\}$ is NOT the graph of a function.

(2). There IS a function $f(x)$ for $|x|<1$ such that $f(x)$ is differentiable and $x^2+f(x)^2=1.$ In fact there are 2 such functions. And we have $$0= \frac {d}{ dx} (1) =\frac {d}{ dx} (x^2+f(x)^2)=2x + \frac {d}{ dx }(f(x)^2)=2x+2f(x)f'(x).$$

Note: In differentiation, it matters what you differentiate BY. Just as in division, where it matters whether you divide by x or by y. The derivative of $y^2$ with respect to $y$ is $2y.$ The derivative of $y^2$ with respect to $x$ is $2y \frac {dy}{dx}.$


I suppose that the use of the implicit function theorem could make things clearer.

Consider $$F(x,y)=0$$ and compute $F'_x(x,y)$ considering $y$ as a constant and then $F'_y(x,y)$ considering $x$ as a constant. Now, the implicit function theorem $$\frac{dy}{dx}=-\frac{F'_x(x,y) }{F'_y(x,y) }$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.