Dave's Math Tables: Basic Derivative Identities
(Math | Calculus | Derivatives | Identities | Basic)
(d/dx) c f(x) = c (d/dx) f(x)
(d/dx) [ f(x) + g(x) ] = (d/dx)f(x) + (d/dx)g(x)
(d/dx) f(g(x)) = (d/du) f(u)  (d/dx) g(x)


Proof of (d/dx) c f(x) = c (d/dx) f(x) : from the definition

Given: (d/dx) f(x) = lim(d->0) ( f(x+d)-f(x) )/d
Solve:

(d/dx) c f(x) = lim(d->0) (c f(x+d)) - (c f(x))/d = c * (f(x+d) - f(x))/d = c * (d/dx) f(x)

Proof of (d/dx) (f(x) + g(x)) = (d/dx) f(x) + (d/dx) g(x) : from the definition

Given: (d/dx) f(x) = lim(d->0) ( f(x+d)-f(x) )/d
Solve:

(d/dx) (f(x) + g(x)) = lim(d->0) [ (f(x+d) + g(x+d)) - (f(x) + g(x)) ] / d
= (f(x+d)-f(x))/d + (g(x+d)-g(x))/d = (d/dx) f(x) + (d/dx) g(x)

Proof of Chain Rule : (d/dx) f(g(x)) = (d/dg) f(g) (d/dx) g(x) : from the definition

Given: (d/dx) f(x) = lim(d->0) ( f(x+d)-f(x) )/d
Solve:

(d/dx) f(g(x)) = df/dx = (f(g(x+d) - f(g(x))/d
df/dx * 1/(dg/dx) = [ (f(g(x+d) - f(g(x))/d ] * [ d/(g(x+d) - g(x)) ]
= ( f(g(x+d))-f(g(x)) )/(g(x+d)-g(x)) = df/dg
df/dx = df/dg * dg/dx