2

If we have an equation like

y = x^2

This implies that

y’ = 2x

If we have an equation like

x = 4x^2

and we take the derivative of both sides we get

1 = 8x

With the solution x = 1/8, which is not the solution to the original equation. This is instead the value where the slopes of both sides of the equation are equal.

So when (and why) is an equation resulting from the derivation of both sides of some original equation implied by the original equation.

Frasch
  • 347

2 Answers2

4

Your problem comes from thinking that you take derivatives of equations. But you don't. You take derivatives of functions.

For example, you can think of the equation $$ x^2 -1 = (x-1)(x+1) $$ as telling you two different ways to write the same function of $x$. When you differentiate the two expressions using the rules for derivatives you will get the same answer, in two different forms.

The equation in your question, $$ x = 4x^2 , $$ does not say that two functions are equal. It asks for the value of $x$ at which those two functions happen to have the same value. That's just a number. It makes no sense to take the derivative.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • 1
    In other words, if $f(x)=x$ and $g(x)=4x^2$, then $f(x)=g(x)$ is an equation, not an identity. – Steven Alexis Gregory May 26 '19 at 14:04
  • 1
    @stevengregory With those meanings for $f$ and $g$ the equation "$f=g$" is false. They are different functions. This may help: https://math.stackexchange.com/questions/2738360/what-exactly-is-an-equation/2738382#2738382 – Ethan Bolker May 26 '19 at 14:10
  • So we can take the derivative of both sides of an identity but not an equation? – Frasch May 26 '19 at 14:11
  • 2
    @FrankieScheuer You can say that but I would not. Skip the words "identity" and "equation" and "derivative of both sides". Just think about differentiating functions. An identity tells you that you have two different ways to write the same function. See the link in an earlier comment. – Ethan Bolker May 26 '19 at 14:32
  • What would something like y = x^2 be? x^2 is a function and I understand that I can take the derivative of it, but to get y’ = 2x aren’t we taking the derivative of both sides? Would you say that y = x^2 is an identity, and taking the derivative of both sides is really just differentiating a function because both sides are the same function y? – Frasch May 26 '19 at 15:03
  • @EthanBolker Yes. Except there is nothing wrong with saying "solve $f(x)=g(x)$ for $x$. Context is very important. – Steven Alexis Gregory May 26 '19 at 15:03
  • 1
    @FrankieScheuer The equation $y = x^2$ is implicitly defining how $y$ is a function of $x$. So it expresses the equality of two ways to write the same function. That's why $y' = 2x$. Please don't think about "taking the derivative of both sides". – Ethan Bolker May 26 '19 at 15:34
  • What about implicit differentiation? I don’t know how to think of that in any way other than taking the derivative of both sides. – Frasch May 26 '19 at 16:05
  • 1
    @FrankieScheuer If, say, $\sin(xy) = xy^2$ and $y$ depends (implicitly) on $x$ then each side of that equation defines the same function of $x$, so has the same derivative, which will involve the derivative of $y$ (using the chain rule). Operationally this is "differentiating both sides" but conceptially not. If you keep the concept in mind you won't make the kind of mistake that led you to ask this (interesting) question. – Ethan Bolker May 26 '19 at 18:17
-1

The solutions of $f(x)=g(x)$ are the points of intersection of $f$ and $g$; the solutions of $f'(x)=g'(x)$ are the points where $f$ and $g$ have parallel tangents.

In our case $f(x)=x$ and $g(x)=4x^2$. The meet for $x\in\{0,1/4\}$. Now $f'(x)=g'(x)\iff x=1/8$, for which $x$ the functions are parallel (but not meet).

Michael Hoppe
  • 18,103
  • 3
  • 32
  • 49