Derivatives
Using the
Power Rule
4.1
Find the derivative using the definition.
f  x  h  f  x
dy
 lim
dx h0
h
y  x  2x  3
2

  x  h  2  2  x  h   3  x 2  2 x  3

 lim 
h 0
h

x 2  2 xh  h 2  2 x  2h  3  x 2  2 x  3
 lim
h 0
h
2 xh  h 2  2h
 lim
h 0
h
h  2 x  h  2   lim 2 x  h  2


 lim
h

0
h 0
h
 2x  2

Finally… the shortcut!!!!!!!
Same problem…
Find the derivative using the power rule.
y  x  2x  3
2

2
dy
d y
Find
and
.
2
dx
dx
y  3x  5x  7 x  4
3
2

Find the derivative using the power rule.
1
2
3
y  4  3 5 x 3 x
2x
x

P200 #3-17 odd
Check your answers!
Write the equation of the tangent at the given point.
f ( x)  x  5x  2 at P  (1, 2)
3
How could you use your graphing calculator to answer
this question more quickly?

Determine the values of x for which the function has a
horizontal tangent. Verify graphically.
y  x  2x  2
4
2
How would you change this process if you were asked
to find where the function has a tangent that is parallel
to y  3x  4 ?

REVIEW
Given that f(x) is a position function
relating distance in terms of time.
f ( x)  position (distance and time)
f ( x)  instantaneous velocity
f ( x)  acceleration
d
average velocity =
t
A particles moves along a coordinate line according to
the position function below.
s(t )  t  4t  3
3
a) Determine the average velocity of the particle from
t=2 to t=4.
b) Determine the instantaneous velocity of the particle
at t=2.
c) At what time(s) is the particle at rest?
d) When is the particle moving to the left?

P200 #1-31odd, 26a-c,30