Why can I take the derivative of both sides of the equation? Proof of the power rule for rational numbers I am trying to understand the proof for the power rule for rational numbers. Please take into consideration, I am a newbie to calculus. I am just trying to learn it from a book. (Calculus the easy way, Douglas Downing) I can't seem to understand why I can differentiate both sides of the equation. I read that it is implicite differentiation. I watched some youtube videos, but I dont understand this implicit thing.
The proof starts like this:
$y = x^{p\over q}$
$y^q = x^p$
${d\over dx}y^q={d\over dx} x^p $
 A: Implicit differentiation is an application of the chain rule to find a derivative of a function that is "implicitly defined": Instead of having $y=f(x)$, we have an equation $F(x,y)=0$ that may (or may not) define $y$ implicitly as a differentiable function of $x$.
For example, if we have $F(x,y) = x^2+y^2 - 1 = 0$, we assume that this equation defines $y$ as a differentiable function of $x$ on some interval. If we say $y=\phi(x)$, then
$$F(x,\phi(x)) = x^2 + \phi(x)^2 - 1 = 0.$$
We differentiate this function of $x$ to get
$$2x + 2\phi(x)\phi'(x) = 0, \quad\text{so}\quad \phi'(x) = -\frac{x}{\phi(x)}.$$
More casually, we skip the step of writing the explicit function $\phi$ and write
$$2x + 2y\frac{dy}{dx} = 0, \quad\text{and so}\quad \frac{dy}{dx} = -\frac xy.$$
(Notice that you must be at points $(x,y)$ with $y\ne 0$ for this to make sense.)
Indeed, in this case, we can solve explicitly for $y=\sqrt{1-x^2}$ if we are on the upper semicircle and for $y=-\sqrt{1-x^2}$ if we are on the lower semicircle. At the points $(\pm 1,0)$ the two semicircles join and we do not have $y$ nearby as a function — let alone a differentiable function — of $x$. Also, note that if we differentiate $y=\sqrt{1-x^2}$, we do indeed get
$$\frac{dy}{dx} = \frac{-2x}{2\sqrt{1-x^2}} = -\frac xy,$$
as before. (You can check that it works out similarly for $y=-\sqrt{1-x^2}$.)
In the problem you posed, you get $F(x,y) = y^q - x^p = 0$. Assuming that $y$ is a differentiable function $y=\phi(x)$, we proceed to differentiate:
$$q\phi(x)^{q-1}\phi'(x) = px^{p-1}, \quad\text{so}\quad \phi'(x) = \frac{px^{p-1}}{q\phi(x)^{q-1}}.$$
We can then simplify, since $\phi(x) = x^{p/q}$, to obtain
$$\frac{dy}{dx}=\phi'(x)=\frac pq \frac{x^{p-1}}{(x^{p/q})^{q-1}} = \frac pq x^{p-1-\frac pq(q-1)} = \frac pq x^{\frac pq - 1}.$$
Again, I reiterate that the algorithm of implicit differentiate proceeds by assuming that $y$ is defined (on some interval) by a differentiable function of $x$ and then applying product rule, quotient rule, sum rule, and chain rule. The proof that this assumption is valid under certain circumstances comes from a theorem called the Implicit Function Theorem, which is proved in more advanced mathematics courses.
A: Write $y=h(x)$ , then it becomes apparent.
It seems quite trivial that: 
If for all $x$, $f(x)=g(x)$, 
then, for all $x$ where at least one of g or f's the derivative exists, $\dfrac{d}{dx} f(x)=  \dfrac{d}{dx} g(x)$

Proof?
For all $x : \;\;$ $f(x)=g(x)$ $\implies$ For all $x, h \;:\;\;$ $f(x+h)=g(x+h)$
Thus, $$\dfrac{f(x+h)-f(x)}{h} = \dfrac{g(x+h)-g(x)}{h}$$
$$\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \dfrac{g(x+h)-g(x)}{h}$$
If one of those limits exists, then the other exists too (by algebra of limits); and hence $f'=g'$

Note @TedShrifrin's comments below. 
What they said, is that $assumptions$ are important. 
In the above proof for example, the assumption is that at least one of
the two limits exists, after which we can write $f'=g'$
Now, it is indeed true without a doubt that $d/dx(y^q)=d/dx(x^p)$.
The natural next step would be to use chain rule to say $d/dx(y^q)=qy^{q-1}(dy/dx)$ 
However, for this equality to be true, we need to assume the existence of $dy/dx$ - that is, we need to assume that $y$ is a differentiable function of $x$.
