The Product Rule · The Product Rule We’re going to nd the derivative of products of functions,...

The Product Rule

The Product Rule

We’re going to find the derivative of products of functions, forexample:

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

f (x) = u(x)v(x)

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

f (x) = u(x)v(x)

We apply the definition:

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

f (x) = u(x)v(x)

We apply the definition:

f ′(x) = lim∆x→0

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

f (x) = u(x)v(x)

We apply the definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

f (x) = u(x)v(x)

We apply the definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x

= lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

f (x) = u(x)v(x)

We apply the definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x

= lim∆x→0

������

�����: f (x+∆x)

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

f (x) = u(x)v(x)

We apply the definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x

= lim∆x→0

u(x + ∆x)v(x + ∆x) −�����: f (x)

u(x)v(x)

∆x

The Product Rule

We’re going to find the derivative of products of functions, forexample:

f (x) = x sin x

Let’s consider a function:

f (x) = u(x)v(x)

We apply the definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x

= lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

The Product Rule

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

Now we apply an algebraic trick: add and subtract u(x)v(x + ∆x).

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

Now we apply an algebraic trick: add and subtract u(x)v(x + ∆x).

u(x + ∆x)v(x + ∆x) − u(x)v(x + ∆x) + u(x)v(x + ∆x) − u(x)v(x)

∆x

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

Now we apply an algebraic trick: add and subtract u(x)v(x + ∆x).

u(x + ∆x)v(x + ∆x) − u(x)v(x + ∆x) + u(x)v(x + ∆x) − u(x)v(x)

∆x

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

Now we apply an algebraic trick: add and subtract u(x)v(x + ∆x).

u(x + ∆x)v(x + ∆x) − u(x)v(x + ∆x) + u(x)v(x + ∆x) − u(x)v(x)

∆x

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

Now we apply an algebraic trick: add and subtract u(x)v(x + ∆x).

u(x + ∆x)v(x + ∆x) − u(x)v(x + ∆x) + u(x)v(x + ∆x) − u(x)v(x)

∆x

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

Now we apply an algebraic trick: add and subtract u(x)v(x + ∆x).

u(x + ∆x)v(x + ∆x) − u(x)v(x + ∆x) + u(x)v(x + ∆x) − u(x)v(x)

∆x

= lim∆x→0

v(x + ∆x) [u(x + ∆x) − u(x)] + u(x) [v(x + ∆x) − v(x)]

∆x

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

Now we apply an algebraic trick: add and subtract u(x)v(x + ∆x).

u(x + ∆x)v(x + ∆x) − u(x)v(x + ∆x) + u(x)v(x + ∆x) − u(x)v(x)

∆x

= lim∆x→0

v(x + ∆x) [u(x + ∆x) − u(x)] + u(x) [v(x + ∆x) − v(x)]

∆x

= lim∆x→0

v(x + ∆x)u(x + ∆x) − u(x)

∆x+

The Product Rule

f ′(x) = lim∆x→0

u(x + ∆x)v(x + ∆x) − u(x)v(x)

∆x

Now we apply an algebraic trick: add and subtract u(x)v(x + ∆x).

u(x + ∆x)v(x + ∆x) − u(x)v(x + ∆x) + u(x)v(x + ∆x) − u(x)v(x)

∆x

= lim∆x→0

v(x + ∆x) [u(x + ∆x) − u(x)] + u(x) [v(x + ∆x) − v(x)]

∆x

= lim∆x→0

v(x+∆x)u(x + ∆x) − u(x)

∆x+ lim

∆x→0u(x)

v(x + ∆x) − v(x)

∆x

The Product Rule

The Product Rule

= lim∆x→0

v(x+∆x)u(x + ∆x) − u(x)

∆x+ lim

∆x→0u(x)

v(x + ∆x) − v(x)

∆x

The Product Rule

= lim∆x→0

v(x+∆x)u(x + ∆x) − u(x)

∆x+ lim

∆x→0u(x)

v(x + ∆x) − v(x)

∆x

= lim∆x→0

v(x + ∆x). lim∆x→0

u(x + ∆x) − u(x)

∆x+

The Product Rule

= lim∆x→0

v(x+∆x)u(x + ∆x) − u(x)

∆x+ lim

∆x→0u(x)

v(x + ∆x) − v(x)

∆x

=��

������

�:v(x)lim

∆x→0v(x + ∆x). lim

∆x→0

u(x + ∆x) − u(x)

∆x+

The Product Rule

= lim∆x→0

v(x+∆x)u(x + ∆x) − u(x)

∆x+ lim

∆x→0u(x)

v(x + ∆x) − v(x)

∆x

= lim∆x→0

v(x + ∆x).���

�����

����:u′(x)

lim∆x→0

u(x + ∆x) − u(x)

∆x+

The Product Rule

= lim∆x→0

v(x+∆x)u(x + ∆x) − u(x)

∆x+ lim

∆x→0u(x)

v(x + ∆x) − v(x)

∆x

= lim∆x→0

v(x + ∆x).���

�����

����:u′(x)

lim∆x→0

u(x + ∆x) − u(x)

∆x+

+u(x) lim∆x→0

v(x + ∆x) − v(x)

∆x

The Product Rule

= lim∆x→0

v(x+∆x)u(x + ∆x) − u(x)

∆x+ lim

∆x→0u(x)

v(x + ∆x) − v(x)

∆x

= lim∆x→0

v(x + ∆x).���

�����

����:u′(x)

lim∆x→0

u(x + ∆x) − u(x)

∆x+

+u(x)���

������

���:v ′(x)

lim∆x→0

v(x + ∆x) − v(x)

∆x

The Product Rule

= lim∆x→0

v(x+∆x)u(x + ∆x) − u(x)

∆x+ lim

∆x→0u(x)

v(x + ∆x) − v(x)

∆x

= lim∆x→0

v(x + ∆x).���

�����

����:u′(x)

lim∆x→0

u(x + ∆x) − u(x)

∆x+

+u(x)���

������

���:v ′(x)

lim∆x→0

v(x + ∆x) − v(x)

∆x

= v(x)u′(x) + u(x)v ′(x)

The Product Rule

The Product Rule

That is:

The Product Rule

That is:

(u(x)v(x))′ = u′(x)v(x) + u(x)v ′(x)

The Product Rule

That is:

(u(x)v(x))′ = u′(x)v(x) + u(x)v ′(x)

The Product Rule

That is:

(u(x)v(x))′ = u′(x)v(x) + u(x)v ′(x)

This is the product rule!