i was fiddling around with implicit differentiation the other day.

d (x^n) can be rewritten as
n / b * x ^ (n-b) * d(x^b),
where b is constant

so if b = 1, n = 2

d (x^2)
= x ^ (2-1) * d(2*x^1/1)
= x * d(2x)
= 2x * d(x)

but, what if b = 0

the RHS expression then becomes

n/0 * x^n d(x^0)

then d(x^0) = 0, since x^0 is a constant 1

i.e

d(x^n) = n*x^n * (0 / 0)

but

d(x^n) / dx = n*x^(n-1)

the RHS is n*x^n * (0/0)

i.e

n*x^n * (0/0) / dx = n*x^(n-1)

i.e

(0/0) / dx = 1 / x

dx / x = ? (0 / 0)?