d/dx[x^n]
= n*x^(n-1)
power rule
d/dx[c]
= 0
derivative of a constant
d/dx[f(x) +/- g(x)]
= f’(x) +/- g’(x)
sum and difference rule
d/dx[c*f(x)]
= cd/dx[f(x)] = cf’(x)
d/dx[a^x]
a^x*ln(a)
d/dx[e^x]
= e^x
d/dx[loga(x)]
= 1/(x*ln(a))
d/dx[ln(x)]
= 1/x
d/dx[sin(x)]
= cos(x)
d/dx[cos(x)]
= -sin(x)
d/dx[tan(x)]
= sec^2(x) = [sec(x)]^2
d/dx[csc(x)]
= -csc(x)*cot(x)
d/dx[sec(x)]
= sec(x)*tan(x)
d/dx[cot(x)]
= -csc^2(x) = -[csc(x)]^2
product rule
h(x)=f(x)g(x)
h’(x)=f’(x)g(x)+f(x)*g’(x)
chain rule
y=f(u(x))
y’=f’(u(x))*u’(x)
quotient rule
y=f(x)/g(x)
dy=f’(x)g(x)-f(x)g’(x)
dx [g(x)]^2
L’Hopital’s rule
let f(x) and g(x) be differentiable functions
either lim f(x)–>0 or–> +/-infinity
x->a g(x)
then apply L’Hopital’s rule:
lim f(x) = lim f’(x) = lim f”(x)
x->a g(x) x->a g’(x) x->a g”(x)
as long 2nd derivative does not approach 0/0 or infinity
Mean Value Theorem (MVT)
if f(x) is continuous on [a,b] and differentiable on (a,b)…
then there MUST exist at least one value, c, on (a,b) such that…
f’(c) = [f(b)-f(a)]/[b-a]
