# B Equation vs. Identity

#### FAS1998

Why are we allowed to differentiate both sides of something like

$y=x^2$

but not something like

$x=x^2$

I believe the answer might be that the first equation is an identity that is true for all values while the second equation is an equation and is only true for some values.

Although I’m not entirely sure if this is correct, and if it is correct, I’m not entirely sure how to distinguish between identities and equations.

#### jedishrfu

Mentor
Who said you can’t differentiate on both sides?

#### FAS1998

Who said you can’t differentiate on both sides?
The solution to the second equation is x = 0 or x = 1.

If you differentiate both sides you get

$1=2x$

Which has the solution x = 1/2

#### fresh_42

Mentor
2018 Award
FAS1998 said:
Why are we allowed to differentiate both sides of something like
$y=x^2$
but not something like
$x=x^2$
fresh_42 said:
And it says: $x=x^2 \Longrightarrow \dfrac{d}{dx}\,x=\dfrac{d}{dx}\,x^2 \Longrightarrow 1=2x \Longrightarrow x=\dfrac{1}{2}$

which is the point at which both functions have the same slope $1$.
The equation $x = x^2$ doesn't represent a function, whereas the first equation above does represent a function.

Differentiating an conditional equation such as $x = x^2$ doesn't completely make sense, because this isn't a function, and functions are the inputs to the differentiation operator. However, if you consider each side as representing its own function; i.e., as $y_1 = x$ and $y_2 = x^2$, then differentiating each of these equations and setting $y_1' = y_2'$ does result in a solution $x = 1/2$.

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#### jedishrfu

Mentor
Also $y=x^2$ is not an identity, it’s a function of the form y=f(x) over some domain assumed to be reals but could be whatever is meaningful, like integers or complex numbers...

#### FAS1998

And it says: $x=x^2 \Longrightarrow \dfrac{d}{dx}\,x=\dfrac{d}{dx}\,x^2 \Longrightarrow 1=2x \Longrightarrow x=\dfrac{1}{2}$
which is the point at which both functions have the same slope $1$.
I don’t quite understand what’s going on there.

I understand that you’re taking the derivative of both sides of the equation, so the solution of the new equation will be the values of x where the derivatives are equal.

But I don’t see how it makes sense to say that

$x=x^2 \Longrightarrow ... \Longrightarrow x=\dfrac{1}{2}$

For the first equation to imply that x = 1/2, wouldn’t x = 1/2 have to be a solution to the first equation?

#### FAS1998

Also $y=x^2$ is not an identity, it’s a function of the form y=f(x) over some domain assumed to be reals but could be whatever is meaningful, like integers or complex numbers...
So would the symbol y in this context represent a different kind of thing than the variable x? Or can y be both a variable and a function of x?

#### jedishrfu

Mentor
Think of it as two functions x and x^2 and you differentiate to find where they both have the same slope.

#### fresh_42

Mentor
2018 Award
But I don’t see how it makes sense to say that

$x=x^2 \Longrightarrow ... \Longrightarrow x=\dfrac{1}{2}$

For the first equation to imply that $x = 1/2$, wouldn’t $x = 1/2$ have to be a solution to the first equation?
No, since differentiation is one way. If we want to differentiate $x$ or $x^2$ we assume that they are functions $x \stackrel{f}{\longmapsto} x\, , \,x\stackrel{g}{\longmapsto} x^2$ resp. Now differentiation is - if written accurately - the following operation:
\begin{align*}
x&=x^2 \text{ interpreted as } f(x) = g(x) \qquad \left| \qquad \left. \dfrac{d}{dx}\right|_{x=p} \right.\\
\left. \dfrac{d}{dx}\right|_{x=p} \,x &= \left. \dfrac{d}{dx}\right|_{x=p}\, x^2\\
p &=2p\\
p &=\dfrac{1}{2}
\end{align*}
We evaluate the differential at a certain point $p$, and that is the slope at this point of $f(x)\, , \,g(x)$ resp. It is a mistake to omit this notation, although it is done all the time. Without the evaluation, we get
$$1=\dfrac{d}{dx}\,x =\dfrac{d}{dx}\,f=\dfrac{d}{dx}\,g= \dfrac{d}{dx}\,x^2 = 2x$$
which is an equation of two functions, namely $x \longmapsto f'(x)$ and $x \longmapsto g'(x)$; or $p \longmapsto f'(p)$ and $p \longmapsto g'(p)$ to emphasise the new role of $x$. To set them equal is as usual the same as to ask where those functions intersect, which they do at $p=\dfrac{1}{2}.$ But they are two different functions than those we started with.

The confusion is only due to the fact, that those functions nor the evaluation point are mentioned if we only write $x$ or $x^2$.

#### suremarc

I prefer to think of it this way: you cannot differentiate $x=x^2$ because it does not have the structure of a differentiable curve. The solution set consists of two isolated points; thus, we cannot define the derivative on the solution set.

On the other hand, $y=x^2$ has the structure of a curve, so we can ask about the derivative which turns out to be well-defined.

#### PeroK

Homework Helper
Gold Member
2018 Award
Why are we allowed to differentiate both sides of something like

$y=x^2$

but not something like

$x=x^2$

I believe the answer might be that the first equation is an identity that is true for all values while the second equation is an equation and is only true for some values.

Although I’m not entirely sure if this is correct, and if it is correct, I’m not entirely sure how to distinguish between identities and equations.
In my view this depends on the mathematical conventions of what we mean by such equations.

If you write: Let $y = x^2$, then that is actually ambiguous. It could mean:

Let $y$ be a function of $x$ defined by $y = x^2$.

In which case, you are free to differentiate this function.

Or, it could mean:

Let $y_0$ be a specific number that satisfies $y_0 = x_0^2$.

In this case, technically, there is no function to differentiate. Although, you could easily turn this into a function by considering $x_0$ as a variable and $y_0$ as a function of this variable.

If, however, you write: Let $x = x^2$, then in my opinion this can only mean that you have a specific number that satifies $x_0 = x_0^2$. You could, in fact, rewrite this as:

$x^2 - x = 0$

In this case, there is nothing to differentiate, as all you have is at most two solutions to this quadratic equation.

In summary, I would compare your original question to the following:

$y = x^2 - x$

Is the definition of a quadratic function, which can be differentiated.

$x^2 - x = 0$

Is an equation for the (discrete) roots of a quadratic function and cannot be differentiated (because the equality only holds for at most two values of $x$).

Note, finally, that an identity is something like:

$\cos^2x + \sin^2 x = 1$

That is an equation that holds for all points. I.e. the graph of the function is the line $y = 1$. So, that can be differentiated.

Note that, to be very precise here, the symbol $1$ actually stands for the function that is identically $1$, which is why it can be differentiated.

"Equation vs. Identity"

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