Can we differentiate equations without changing the solutions? Suppose we have $x^2+4x+3=0$ 
We differentiate both sides to get $2x+4=0$. 
So $x=-2$ 
But $(-2)^2+4(-2)+3=0$ 

$4-8+3=-1$ is not equal to $0$. 
Which tells that we cannot differentiate equalities. But in many proofs, like in integral calculus, differentiating both sides occurs. So, where is the fallacy?
 A: In a word, quantifiers.
You have to distinguish between equations that hold for all $x$ and those that hold for a particular $x$.  An equation that is true for all $x$ in an interval (and where both sides are differentiable) remains true when differentiated.  One that is true for a particular $x$ is not likely to remain true.
A: Consider functions $f(x)=g(x)$ take any boundaries you like.
Our statement here tells us that, a function f(x) is same as or equal to g(x)
The fallacy is that differentiation is an operator. Here's what it means.
If you are differentiating both side of that equation you would get f$\;$'(x)=g'(x)
this statement reads us that(one of the meaning) the rate of change of g(x) is same as or equal to the rate of change of f(x)
Now even though equality or functions were same. Both of them has very different meaning , especially when equality doesn't hold for all x. for e.g, in your case when.
$x^2+4x+3=0$ here, $f(x)=x^2+4x+3$ and $g(x)=0$
Where equality holds for $x=3$ or $x=1$.
Now where their explicit values might be equal their rates may or may not be equal at that same number.
i.e., $f'(x)=g'(x)$ , may or may not be true for the same x for which $f(x)=g(x)$
A: Term by term differentiation of an infinite series identity
$$ \sinh(x) =  \Sigma  \dfrac{x^{2 n -1}} { (2 n-1)!}$$
leads to
$$ \cosh(x) =  \Sigma  \dfrac{x^{2 n }} { (2 n)!}$$
So for all identities it seems to be allowable, but not for functions in general.
