Derivatives
LRT
02/08/2017
Two basic rules
The goal today is to introduce some techniques for
calculating derivatives so that we do not always have to go
back to the definition.
d d 1. (Constant multiple)
cf (x) = c
f (x) for any
dx
dx
real number c.
2. (Sum)
d d d f (x) ± g(x) =
f (x) ±
g(x) .
dx
dx
dx
d d 1. (Constant multiple)
cf (x) = c
f (x) for any
dx
dx
real number c.
2. (Sum)
d d d f (x) ± g(x) =
f (x) ±
g(x) .
dx
dx
dx
We can combine 1 and 2 to get that if
g(x) = c1 f1 (x) ± c2 f2 (x) ± · · · ± cr fr (x)
then
g 0 (x) = c1 f10 (x) ± c2 f20 (x) ± · · · ± cr fr0 (x)
Two examples
1. (Constant)
d
(c) = 0 for any real number c.
dx
d r
(x ) = r xr−1 for any real number r and
dx
any x for which xr−1 is defined.
2. (Power)
Examples 1 and 2 tell us how to differentiate specific
families of functions, constant and power.
Rules 1 and 2 tell us how to differentiate algebraic
combinations of functions if we know how the differentiate
the individual functions.
Examples
Differentiate f (x) = 3x5 − 5x2 + 2.
f 0 (x) = 3 · 5x5−1 − 5 · 2x2−1 + 0 = 15x4 − 10x.
Differentiate g(t) = 3t5 − 5t2 + 2.
g 0 (t) = 15t4 − 10t since the functions g and f are the same.
If y = 3x5 − 5x2 + 2 find
dy
= 15x4 − 10x.
dx
dy
.
dx
Find g 0 (t) if g(t) = 31 t3 − 23 t2 .
g 0 (t) =
1
3
· 3 t3−1 − 23 · 2 t2−1 = t2 − 3t.
u2 − 3u + 4
.
7
1 d (u2 − 3u + 4)
1
0
h (u) =
= 2u − 3 .
7
du
7
Find D2 h(u) if h(u) =
Hence D2 h(u) = h0 (2) =
1
1
2·2−3 = .
7
7
Find an equation of the tangent line to the graph of
y = 3x5 − 5x2 + 2 when x = 1.
New Notation
dy 4
= (15x − 10x)
=
dx x=1
x=1
15 − 10 = 5.
Hence the slope is 5 and the point with x = 1 and
y = 3 · 15 − 5 · 14 + 2 = 0, or (1, 0), is on the tangent line.
y − 0 = 5(x − 1)
so y = 5x − 5.
Here is the graph plus the tangent line at x = 1.
2
1
-3
-2
-1
0
-1
-2
1
2
3
Let f (x) = x0.8 + x3 + π 4 . Find f 0 (x).
f 0 (x) = 0.8x−0.2 + 3x2 + 0.
NOT f 0 (x) = 0.8x−0.2 + 3x2 + 4π 4−1
Let f (x) =
x3 − 4x2 + 8
. Find f 0 (x).
x3
As written our rules fail to apply.
4x2
8
BUT f (x) = 1 − 3 + 3 = 1 − 4x−1 + 8x−3 .
x
x
Then f 0 (x) = 0 − 4(−1)x−1−1 + 8(−3)x−3−1 = 4x−2 − 24x−4
1
d Let f (y) = √ . Find f 0 (y) =
f (y) .
y
dy
1
1
1
f (y) = √ = 1 = y − 2 so
y
y2
1
1
3
f 0 (y) = − 21 y − 2 −1 = − 12 y − 2 = − p .
2 y3
An example which comes up a lot.
If f (x) =
√
√
x, find f 0 (x).
1
1
1
1
x = x 2 so f 0 (x) = 21 x 2 −1 = 12 x− 2 = √
2 x
Let f (u) =
√
3
1
u4 . Find f 0 (u).
4
f (u) = (u4 ) 3 = u 3 .
4
1
f 0 (u) = 43 u 3 −1 = 43 u 3
Let f (x) = (x − 1)2 + x + 8. Find f 0 (x).
f (x) = x2 − 2x + 1 + x + 8 = x2 − x + 9
so
f 0 (x) = 2x − 1 + 0 = 2x − 1.
Right answer - wrong method:
f 0 (x) = 2(x − 1) + 1 = 2x − 1.
Wrong answer - wrong method: Change f to
g(x) = (2x − 1)2 + x + 8 = 4x2 − 4x + 1 + x + 8 = 4x2 − 3x + 9
so g 0 (x) = 8x − 3.
Which is NOT 2(2x − 1) + 1 = 4x − 1.
Eventually we will learn a formula for
answer is NOT always 2f (x).
2
d f (x) but the
dx
#64 in book
The total groundfish population on Georges Bank off New
England between 1989 and 1999 is approximated by the
function
f (t) = 5.303t2 − 53.977t + 253.8 (0 6 t 6 10)
where f (t) is measured in thousands of metric tons and t in
years, with t = 0 corresponding to the beginning year of
1989.
a. What is the rate of change of the groundfish population
at the beginning of 1994? At the beginning of 1996?
f 0 (t) = 5.303 · 2t − 53.977 + 0 = 10.606t − 53.997.
1994 is t = 5 so the rate of change is −0.967.
1996 is t = 7 so the rate of change is 20.245.
b. Fishing restrictions were imposed on Dec. 7, 1994. Were
the conservation measures effective?
In 1994 the fish population was declining whereas by 1996
it was increasing so things were certainly getting better for
the groundfish population. Whether this turnaround is due
to the imposed conservation measures or not is a question
for biology not mathematics.
There are often several ways to do a√problem.
As an
example, find F 0 (t) if F (t) = 7 t2 − t .
 d 2 √ 

constant multiple rule
7
t − t


 dt
√ F 0 (t) = d
2
7t
−
7
t
basic algebra



dt


 d 2
1

1 − 21

2
7
t
−
t
=
7
2t
−
t


2
 dt
1
1
1
= d
7t2 − 7t 2 = 7 · 2t − 7 · 12 t− 2 = 14t − 72 t− 2



dt

