TITLE: It is possible to apply a derivative to both sides of a given equation and maintain the equivalence of both sides?
QUESTION [13 upvotes]: I have an equation with this shape, where $k,t_1,r \in \Bbb N$:
$$2^{(x+1)^2}=k+t_1(x^2+r)$$
And I noticed that I can find $t_1$ in terms of $x$ as follows:

$$ \frac{{\rm d} }{{\rm d}x} 2^{(x+1)^2} = \frac{{\rm d} }{{\rm d}x} \left( k+t_1(x^2+r) \right)$$

And then I would continue as follows:
$$ 2(x+1)\cdot2^{(x+1)^2}\cdot\ln(2) = t_12x$$
$$ (x+1)\cdot2^{(x+1)^2}\cdot\ln(2) = t_1x$$
So finally:
$$ t_1=\frac{(x+1)\cdot2^{(x+1)^2}\cdot\ln(2)}{x}$$

Is it possible to use a derivative in such case? I think that if both functions $2^{(x+1)^2}$ and $t_1(x^2+r)$ are equivalent and the derivative exists it might be possible, but I am not very sure about the validity of that step. Thank you!

REPLY [2 votes]: What is a derivative? Let $f$ be defined on an open $I \subset \mathbb{R}$ and fix $c \in I$. $f$ is differentiable at $c$ if and only if there exists a real number denoted $f'(c)$ such that $f(x) = f(c) + (x-a)f'(c) + o(x-a)$. It's easy to see, therefore, that the derivative is a "local" property. It matters not only on the value of $f$ at $c$, but at a neighbourhood of $c$. 
What does this have to do with your question, you ask? 
Suppose $f$ and $g$ are differentiable. If $f(a) = g(a)$ for some $a \in I$, then $f'(a) = g'(a)$  needn't be true for any $a \in I$. For example, if you have $x + 1 = 2$, this has a solution $x=1$, but the derivatives of $g(x) := 2$ and $f(x) := x+1$ never coincide. 
On the other hand, if $f(a) = g(a)$ for some $a \in I$ and for each $a$ satisfying $f(a) = g(a)$ there is a $\delta$ such that $f(x) = g(x)$ for all $x \in (a - \delta, a + \delta)$, then we can definitely say $f'(a) = g'(a)$ for all such $a$. 
Of course, if $f$ and $g$ are differentiable and equal everywhere then the above holds clearly and hence $f' = g'$ for all $x \in \mathbb{R}$.