# Derivative Techniques - Jake and Michael

• Power rule

The derivative of x to a power is the power times x to a power one less than the original power.

example 1:

f(x) = x -2 , then f '(x) = -2 x -3 = -2 / x 3

as a short proof:

power rule is true when n = 1:

d
dx
x1

# 1·x0

1;

that it is true when n = 2:

d
dx
x2
= 2x;

and that it is true when n = 3:

d
dx
x3
= 3x².

proof by induction:
The induction hypothesis will be that the power rule is true for n = k:
 d dx xk = k xk−1,
and we must show that it is true for n = k + 1; i.e. that
 d dx xk+1 = (k + 1) xk.
Now,
 d dx xk+1 = d dx x· xk

 d dx xk+1 = x· k xk−1 + xk· 1, according to the product rule,x d dx xk+1 = k xk + xk d dx xk+1 = (k + 1)xk.
Therefore, if the power rule is true for n = k, then it is also true for its succesor, k + 1. And since the rule is true for n = 1, it is therefore true for every natural number.

• Chain Rule

The derivative of a composite function is the derivative of the inside function times the derivative of the outside function (keeping the inside part intact).

• Product Rule

The derivative of two multiplied functions is the addition of the products of one function and the other functions derivative, and vice versa.

Proof:

To prove the product rule, we will express the difference quotient
 simply as Δy Δx . Thus let
y = f g.
Then a change in y -- Δy -- will produce corresponding changes in f andg:
y + Δy = (f + Δf )(g + Δg)
On multiplying out the right-hand side,
y + Δy = f g + f Δg + g Δf + Δf Δg.
 To form Δy Δx , we subtract y -- which is f g -- and divide by Δx: Δy Δx = f Δg Δx + g Δf Δx + Δf Δg Δx
Now let Δx[[image:calc_IMG/Rarrow.gif width="24" height="15"]]0. Hence Δf will approach 0 --

Therefore,
 dy dx = f dg dx + g df dx

example:
f(x) = sinxcosx

f'(x) = (f')(g) + (g')(f)
when,
f = sinx f' = cosxg = cosx g' = -sinx
f'(x) = (cosx)(cosx) + (sinx)(-sinx)
= cos2x - sin2x dx

• Quotient Rule

The derivative of two functions divided is the subtraction of the product of the top function and the bottom function's derivative from the product of the top function's derivative and the bottom function. Then one divides that result by the square of the bottom function.

example:
f(x) = (6x)/(x2 - 4)
f'(x) = h(x) = [(f')(g) - (g')(f)]/(g)2
where,
f = 6x f' = 6g = x2 - 4 g' = 2x
h(x) = [(6)(x2 - 4) - (6x)(2x)]/(x2 - 4)2
= (6x2 - 24) - (12x2)/(x2 - 4)2

= (-6x2 - 24)/(x2 - 4)2

• Deriving an integral function

(d/dx) = f(x)
The derivative of a function in an integral is just the function.

• Logarithmic differentiation
First we consider that the derivative of lnx = 1/x.
Second we consider logarithm rules. Any logarithm with any base can be rewritten via change of base formula...

where b is any constant.

Third we consider that lnx is merely a logarithm with base e.
It follows that we can rewrite any logarithm to contain only ln, which we know how to derive. We will always have an variable ln on top divided by a constant on the bottom.
(d/dx)log_a(x) = (d/dx)lnx/ln(a) = (1/x)/lna = 1/(xlna)

Ex: (d/dx)log_5(x) = (d/dx)lnx/ln(5) = 1/(xln5)

sources:

http://www.themathpage.com/aCalc/rules.htm#power