In books, I often come across situations when they differentiate both sides of an equation or integrate both sides, apply Laplace transform on both sides, etc. Like it's as simple as multiplying both sides or dividing both sides by some constant.

Is it really that simple? It's never explained in the books why are we able to perform those operations. What does differentiating both sides mean? If two things are equal, then their slopes must be equal? While solving, say $x^3+x=7$, we can't just differentiate both sides and solve for $x$.

So, when does it make sense to perform these operations?

  • 1
    $\begingroup$ As an addition to the answer below, any time we have a well defined operation on functions (ex. $\frac{d}{dx}$), that operation preserves equality. This is because such an operation is also a function, only this time it's a function that takes as input functions and spits out other functions (ex. $\frac{df}{dx}(a)$ makes sense). The only outlier here is the indefinite integral, since this operation is only defined up to a constant. $\endgroup$
    – Exit path
    Dec 21, 2017 at 8:03
  • $\begingroup$ Well, you need to be sure that the operation has kernel zero ... $\endgroup$ Dec 21, 2017 at 8:16
  • $\begingroup$ As an aside, if you take the differential of both sides, you get $(3x^2 + 1)\mathrm{d}x = 0$. A true statement, but you should already know $\mathrm{d}x = 0$ on the solution set to $x^3 + x = 7$, so it's not helpful. $\endgroup$
    – user14972
    Dec 21, 2017 at 9:43
  • $\begingroup$ possible duplicate of math.stackexchange.com/questions/1096319/… $\endgroup$ Jun 4, 2018 at 13:23
  • $\begingroup$ possible duplicate of math.stackexchange.com/questions/407822/… $\endgroup$ Jun 4, 2018 at 13:27

1 Answer 1


As mentioned, applying a function or operator to both sides of an equality is fine. It is in recovering the inputs that things get dicey.

In general, undoing a function on both sides of an equation, i.e. hoping that $$ f(x)=f(y)\implies x=y $$ is true only if the original function $f$ is injective. Think of the following $$ x^3=y^3\implies x=y $$ where we took cube roots, the inverse of cubing. Contrast this with, $$ \cos0=\cos2 \pi $$ but $0\ne 2\pi$ because $\arccos$ is not a proper inverse for $\cos$ (it is a multivalued "function" and only works as a proper inverse on restricted intervals), and this is what we applied to both sides to derive the silliness.

However, all is not lost with non injective functions. For example, you are probably familiar with the following sort of argument $$ y^2=x^2\implies|x|=|y| $$ meaning that $x=y$ up to a sign. Similarly, for the previous example
$$ \cos x=\cos y\implies x=y+\pi k,\;k\in \mathbb{Z} $$ So, as long as we know what we may have lost and we keep track of it, it can be possible to "undo" even non injective functions.

Much the same concept is at play when you integrate a differential equation, we know that if we integrate indefinitely and find a primitive, this is really solving for a function up to a constant, for example, $$ x'(t)=c\implies x=ct+d $$ for any $d\in \mathbb{R}$. It really is the same as the square root example, we are keeping track of what was potentially lost in differentiating to get to the formula $x'(t)=c$, namely any information about that constant which vanishes when we differentiate. We essentially "invert" to find a potential class of functions through integration.

More to your examples, you are certainly correct that just because two functions are equal at a point does not mean that their tangents are equal. However, if they are equal in an open neighborhood (and differentiable of course), you may differentiate at some point in the interval and maintain equality. Indeed, in light of the above example (and maybe some visual intuition), you can do this if one is just a vertical shift (i.e. the other plus a constant) of the other in a neighborhood. As to your example, $$ x^3+x=7 $$ is true at points, it is not a functional equation in a neighborhood, so it is not valid to differentiate both sides and claim equality.

As for integral transforms, the story gets a bit more complicated, but the short of it for well behaved functions is that the Fourier transform is indeed invertible, and by the earlier discussion, you may conclude that $$ \hat{f}=\hat{g}\iff f=g $$ for say continuous functions $f$ and $g$. Similarly with the Laplace transform.

I hope this helps! It is a great and thoughtful question.

  • 2
    $\begingroup$ Applying a function to both sides of an equation always maintains the equality, that's basically the definition of a function (clearly $x=y$ implies $x^2=y^2$, but $(-)^2$ is not injective). Your first point has less to do with applying functions and more to do with where the functions are defined. While I understand the spirit of your argument I think this point is likely to confuse $\endgroup$
    – Exit path
    Dec 21, 2017 at 7:57
  • 3
    $\begingroup$ It makes much more sense now :) $\endgroup$
    – Exit path
    Dec 21, 2017 at 8:11

You must log in to answer this question.

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