Dave's Math Tables: Basic Derivative Identities |
(Math | Calculus | Derivatives | Identities | Basic) |
c f(x) = c f(x) [ f(x) + g(x) ] = f(x) + g(x) f(g(x)) = f(u) g(x) |
Given:
f(x) = lim(d->0) ( f(x+d)-f(x) )/d
Solve:
c f(x) = lim(d->0) (c f(x+d)) - (c f(x))/d = c * (f(x+d) - f(x))/d = c * f(x)
Proof of (f(x) + g(x)) = f(x) + g(x) : from the definition
Given:
f(x) = lim(d->0) ( f(x+d)-f(x) )/d
Solve:
(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 = f(x) + g(x)
Proof of Chain Rule : f(g(x)) = f(g) g(x) : from the definition
Given:
f(x) = lim(d->0) ( f(x+d)-f(x) )/d
Solve:
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