How do you find the slope of the tangent line for #f(x) = 3x^2# at (1,3)?

I will assume that you have not yet been taught the rules (shortcuts) for finding derivatives. So, we will use a definition.

The slope of the line tangent to the graph of the function #f# at the point #(a, f(a))# can be defined in several ways (or using several notations). Two of the more common are:

#lim_(xrarra) (f(x)-f(a))/(x-a)# #" "# OR #" "# #lim_(hrarr0) (f(a+h)-f(a))/h#

(Each author,teacher,presenter needs to choose one definition as the 'official' definition. Many will immediately mention other possibilities as 'equivalents'.)

For this question we have #f(x) = 3x^2# and #a=1#

We'll find:

#lim_(xrarra) (f(x)-f(a))/(x-a)#
(Note that substitution gets us the indeterminate form #0/0#. We have some work to do.)

#lim_(xrarr1) (f(x)-f(1))/(x-1) = lim_(xrarr1) (3x^2-3)/(x-1)#

# = lim_(xrarr1) (3(x^2-1))/(x-1)#

# = lim_(xrarr1) (3(x+1)(x-1))/(x-1)#

The expression whose limit we want is equal to #3(x+1)# for all #x# other than #x=1#. The limit doesn't care what happens when #x# is equal to #1#, it wants to onow what happens when #x# is close to #1#, so we get:

#lim_(xrarr1) (3(x+1)(x-1))/(x-1) = lim_(xrarr1) 3(x+1) = 6#

The slope of the tangent we were asked about is #6#.

Short method

For #f(x) = 3x^2#, we get #f'(x) = 3*2 x^(2-1) = 6x#

The slope of the tangent at #1# is #f'(1) = 6(1) =6#