I have seen quite a lot of questions to this effect on the StackExchange already, and have always justified the cancelling of differentials to myself in a rather hand-wavy sort of way - that is, until I was trying to work out some general way to solve an equation of the form:
$$ (f \circ a) (x) - (f \circ b) (x) = c(x) $$
Where I chose a simple $f(u)$, $a(x)$, $b(x)$ and $c(x)$ (like $ f(u) = u^2 $, $ a(x) = x $, $ b(x) = x^2 $ and then $ c(x) = x^2 - x^4 $ ), so as to see whether I could reverse engineer it somehow. My idea was roughly to try and exploit the chain rule by differentiating both sides and to see whether or not it would allow me to factor out some $\frac{df}{dx}$, but when applying the chain rule in the naive, hand-wavy way I mentioned above the cancelling of differentials would collapse the entire equation down to zero. To be more precise, after taking the derivative of
$$ f(x) - f(x^2) = x^2 - x^4 $$
I should have
$$ \frac{df}{da} * 1 - \frac{df}{db} * 2x = 2x - 4x^3 $$
But if I do it purely in terms of $f$, $a$, $b$, and $c$ with the chain rule, I get
$$ \frac{df}{da} * \frac{da}{dx} - \frac{df}{db} * \frac{db}{dx} = \frac{dc}{dx} $$
Which, after cancelling da/da and db/db, the equation is basically
$$ \frac{df}{dx} - \frac{df}{dx} = \frac{dc}{dx} $$ which must be equal to zero ...no?
Long story short, my question is: Have I done something really wrong? If not, what are the subtleties when cancelling differentials? Because right now I'm having a bit of a panic attack thinking that, in general, $\frac{df}{dx} \neq \frac{df}{dx}$ because of the intermediary differentials that may possibly have been cancelled. An answer in layman's terms would be preferred and I apologise in advance for not being able to format the equations properly, as I'm not all that familiar with actively engaging with this forum. Thanks :)
You are using the symbol $$\dfrac{df}{dx}$$ in three different ways and that is what is confusing you. We can write $f$ as a shorthand for $f(x)$ and then $$\dfrac{df}{dx}=\dfrac{df(x)}{dx}=f'(x)$$ Or we can write $f$ as a shorthand for $f(a(x))$ and then $$\dfrac{df}{dx}=\dfrac{df(a(x))}{dx}=f'(a(x))a'(x) $$ Or we can write $f$ as a shorthand for $f(b(x))$ and then $$\dfrac{df}{dx}=\dfrac{df(b(x))}{dx}=f'(b(x))b'(x) $$ If there is any scope for confusion it is best not to write $f$ as it can be ambiguous. Now we can write $$f(a(x)) - f(b(x)) = c(x)$$ and differentiate to get $$f'(a(x))a'(x) - f'(b(x))b(x) = c'(x)$$ and now substitute $f(u) = u^2, a(x) = x, b(x) = x^2 ,c(x) = x^2 - x^4 $ $$2x - 2x^2 2x = 2x-4x^3$$ or, if you prefer, $$\dfrac{df(a(x))}{dx}-\dfrac{df(b(x))}{dx}=\dfrac{dc(x)}{dx}$$ $$\dfrac{df(a(x))}{da(x)}\frac{da(x)}{dx}-\dfrac{df(b(x))}{db(x)}\frac{db(x)}{dx}=\dfrac{dc(x)}{dx}$$ $$2a(x)1-2b(x) 2x=2x-4x^3$$ $$2x-4x^3=2x-4x^3$$