Exercise 1.

Continue this argument to show that the derivative of ^3 is 3^2 for any number . IN MARKUP IN PYTON

()=^1 = /

In this case

()lim0 1/(/+/) = lim0 / -(+)/(+) = lim0 /(+) = /x^2

and we see that the derivative of ^1 is ^2 for any number

Exercise 2.

Continue this argument to show that the derivative of ^2 is 2^3 for any number IN MARKUP IN PYTON

The Product Rule

We will often need to differentiate products like 2sin() and so, in general, we must work out

()()lim0(+)(+)()()/

The key here is to add and subtract terms from the numerator in order to break it into pieces that look like derivatives of and . To wit using red to denote the new stuff,

()()lim0 (+)(+)()()+(+)()(+)()/

=lim0((+)(+)()/+()(+)()/)

=()()+()().

With ()=^2 and ()=sin()we see that the derivative of their product is

^2cos()+2sin()

Exercise 3. Find the derivative of sin()cos() Label and and express to and en route to () IN MARKUP IN PYTON

The Quotient Rule

This same strategy will help us differentiate quotients like sin()/cos()sin()/cos(). In we must work out

(/)()lim0 1/((+)(+)()()) =lim0 1/()(+)(+)()()/(+)

We again add and subtract terms from the numerator in order to break it into pieces that look like derivatives of and . To wit,

(/)()lim0 1/ ()(+)(+)()+()()()()/()(+)

=lim0 (1/(+) (+)()/ ()/()(+)(+)()/)

=()/()()()/^2()

=()()()()/^2()

Hence, with ()=sin() and ()=cos() we find that

tan()= cos()cos()sin()(sin())/cos^2()=1cos^2().