d is just an infinitesimally small delta. So dy/dx is literally just lim (∆ -> 0) ∆y/∆x.
which is the same as lim (x_1 -> x_0) [f(x_0) - f(x_1)] / [x_0 - x_1].
Note: ∆ ->0 isn’t standard notation. But writing ∆x -> 0 requires another step of thinking: y = f(x) therefore ∆y = ∆f(x) = f(x + ∆x) - f(x) so you only need ∆x approaching zero. But I prefer thinking d = lim (∆ -> 0) ∆.
I just think of the definition of a derivative.
dis just an infinitesimally small delta. Sody/dxis literally justlim (∆ -> 0) ∆y/∆x. which is the same aslim (x_1 -> x_0) [f(x_0) - f(x_1)] / [x_0 - x_1].Note:
∆ -> 0isn’t standard notation. But writing∆x -> 0requires another step of thinking:y = f(x)therefore∆y = ∆f(x) = f(x + ∆x) - f(x)so you only need∆xapproaching zero. But I prefer thinkingd = lim (∆ -> 0) ∆.