How do you find the instantaneous rate of change of #g(t)=3t^2+6# at t=4?

Compute the first derivative:

#g'(t) = 6t#

Evaluate it at #t = 4#

#g'(4) = 24#

If you have learned the power rule, constant multiple rule and derivative of a constant, you can quickly find the derivative of #g#.

#g'(t) = 3(2x^1)+0 = 6t#.

To find the instantaneous rate of change at a particular value of #t#, evaluate the derivative at that value of #t#.

At #t=4# the instantaneous rate of change is #g'(4) = 6(4) = 24#.

If you are using a definition then it depends on the particular definition you are using.

There are several ways to express the definition.

One way of expressing it is to give:

The rate of change of #g# with respect to #t# at #t=4# is

#lim_(trarr4)(g(t)-g(4))/(t-4)#.

Another is

The rate of change of #g# with respect to #t# at #t=4# is

#lim_(hrarr0)(g(4+h)-g(4))/h#.

Still another is

The rate of change of #g# with respect to #t# at #t# is

#lim_(hrarr0)(g(t+h)-g(t))/h#.

(After we find this, we evaluate at #t=4#.

Here is the work for the first definition above.

#lim_(trarr4)(g(t)-g(4))/(t-4) =lim_(trarr4) ([3t^2+6]-[3(4)^2+6])/(t-4)# (Observe that is we substitute #t = 4#, we get the indeterminate form #0/0#.)

# = lim_(trarr4) (3t^2+6-48-6)/(t-4)# #" "# (Still #0/0#)

# = lim_(trarr4) (3t^2-48)/(t-4)#

# = lim_(trarr4) (3(t^2-16))/(t-4)#

# = lim_(trarr4) (3(t+4)(t-4)))/(t-4)#

# = lim_(trarr4) 3(t+4)#

# = 3(4+4) = 24#