Lecture Note Differentiation
whent= 3 , p=p(3) 3.1 0.1 3=+×=^24 and so
2
0.1 4 3 1.2
0.24 parts per million per year
0.5 4 17 25
dc
dt
××
===
×+
6 Higher-Order Derivatives
.1 The Second Derivative
the derivative of its derivative. If , the
6
The second derivative of a function is yfx= ()
second derivative is denoted by:
(^) ()
2
2 or
dy
f x
dx
′′
The second derivative gives the rate of change of the rate change of the original
function.
Example 1
Find both the first and second derivatives of the functions:
a. f()xx=^3 − + 12 x 1 b. fx xxx( )= 53342 −−+ 7 c. ()
()
2
32
1
x
fx
x
−
=
−
Example 2
An efficiency study o he mof t rning shift at a certain factory indicates that an average
worker who arrives on the job at 8:00AM. Will have producedQt()=− + +t^32624 t t
units t hours later.
a. Compute the worker’s rate of production at 11:00A.M
b. At what rate is the worker’s rate of production changing
at 11:00
with respect to time
A.M?
rate of production
ctual change in the worker’s rate of production between
Solution
c. Use calculus to estimate the change in the worker’s
between 11:00 and 11:10A.M.
d.Compute the a
11:00 and 11:10A.M.
a. The worker’s rate of production is the first derivative
Q′(t)=−+ + 3122 t^2 t 4
At 11:00 A.M., t= 3 and the rate of production is
Q′() 3 =− × + × + =3 3^2 12 3 24 3 units per hour 3
b. The rate of change of the rate of production is the second derivative
Qt′′()=−+ 612 t
At 11:00 A.M., the rate is
Q′′() 3 =− × + =−6 3 12 6 unit per hour per hour
c. Note that 10 minutes is 1/6 hours, and hence Δt= 16 hour.
Change in rate of production is ()
1
=− × 6 1unit per hour
6
ΔΔQQtt′′′
=−