1.5 Algebraic Rules for Finding Derivatives
The derivative of a function is another function. Suppose we have some function. For
example,
y = x2 = f(x)
Suppose we fix an x and find the rate of change (derivative) at x using the method we
have talked about, i.e.
dy
f(x+h) - f(x)
= f '(x) = lim
dx
h
h0
(x+h)2 - x2
= lim
h
h0
2
x + 2xh +h2 - x2
= lim
h
h0
2
2xh +h
= lim
h
h0
= lim 2x +h
h0
= 2x
Now we let x vary. We get a new
function that assigns to each x the rate of
change at x. This new function is the
derivative function s = f '(x). In our
example
y = f(x) = x2
and
s = f '(x) = 2x
For example, if x = -2, then y = f(-2) = (-2)2 = 4 and s = f '(-2) = (2)(-2) = -4.
As another example, suppose y = 3x = f(x). Then
f(x+h) - f(x)
3(x+h) - 3x
3h
=
lim
=
lim
= lim 3 = 3
h
h
h0
h0
h0 h
h0
s = f '(x) = lim
1.5 - 1
As yet another example, suppose y = 4 = f(x). Then
f(x+h) - f(x)
4-4
=
lim
= lim 0 = 0
h
h0
h0 h
h0
s = f '(x) = lim
In this section we look at algebraic rules for finding the derivative function from
the original function. These rules save us the trouble of finding the derivative by taking a
limit.
dy
Rule #1 (The Linear Function Rule). If y = mx + b, then dx = m.
dy
Examples. If y = 2x + 5. Then dx = 2.
dy
If y = x . Then dx = 1.
dy
If y = 5. Then dx = 0.
Proof of the Linear Function Rule.
dy
[m(x + h) + b] - [mx + b]
mh
= lim h = lim m = m
dx = hlim
h
0
h0
h0
dy
Rule #2 (The Power Rule). If y = xn, then dx = nxn-1.
dy
Examples. If y = x3. Then dx = 3x2.
dy
If y = x = x1. Then dx = x0 = 1.
dy
If y = 1 = x0. Then dx = 0x-1 = 0.
Proof of the Power Rule.
1.5 - 2
dy
f(u) - f(x)
un - xn
=
lim
=
lim
dx u  x u - x
ux u - x
(u - x)(un-1 + un-2x + un-3x2 + un-4x3 + … + uxn-2 + xn-1)
u-x
ux
= lim un-1 + un-2x + un-3x2 + un-4x3 + … + uxn-2 + xn-1
= lim
ux
= nxn-1
Example. Let x = be the length of the side of a cube and V be the volume. One has
V = x3. Suppose currently x = 10 in and V = 1000 in3. Now we increase x to 10.1 in.
Approximatly how much does V increase?
In this case x = 0.1. One has
V dV

= 3x2|x = 10 = 300. So V  (300)(0.1) = 30.
x dx |x = 10
So the volume increses by about 30 in3.
Rule #3 (The Constant Multiple Rule). The derivative of a constant times a function is
the constant times the derivative of the function. More precisely, if y = cf(x), then
dy
dx = cf '(x).
dy
d
Example. If y = 3x2. Then dx = 3 dx [x2] = 3  2x = 6x.
Proof of the Constant Multiple Rule.
dy
cf(x+h) - cf(x)
f(x+h) - f(x)
f(x+h) - f(x)
=
lim
=
lim
c
=
c
lim
= cf '(x)
dx h  0
h
h
h
h0
h0
Rule #4 (The Sum / Difference Rule). The derivative of a sum or difference is the sum
dy
or difference of the derivatives. More precisely, if y = f(x) + g(x), then dx = f '(x) + g'(x)
dy
and if y = f(x) - g(x), then dx = f '(x) - g'(x).
Example. If y = 5x3 + 2x2 - 7x + 3. Then
dy d
d
d
d 3
d 2
3
2
=
[5x
]
+
[2x
]
[7x
3]
=
5
[x
]
+
2
dx dx
dx
dx
dx
dx [x ] – 7
= 15x2 + 4x - 7.
1.5 - 3
Proof of the Sum Rule.
dy
[f(x+h) + g(x+h)] - [f(x) + g(x)]
f(x+h) - f(x) g(x+h) - g(x)
=
lim
=
lim
[
+
]
dx h  0
h
h
h
h0
f(x+h) - f(x)
g(x+h) - g(x)
+
lim
= f '(x) + g'(x)
h
h
h0
h0
= lim
Rule #5 (The Product Rule). The derivative of a product is
(the first)(the derivative of the second) + (the second)(the derivative of the first)
dy
More precisely, if y = f(x)g(x), then dx = f(x)g'(x) + g(x)f '(x).
Example. If y = (5x3 + 7x)(2x2 – 3). Then
dy
d
d
3
2
2
3
=
(5x
+
7x)
[2x
–
3]
+
(2x
–
3)
dx
dx
dx [5x + 7x]
= (5x3 + 7x) 4x + (2x2 – 3) (15x + 7)
If it desired we could multiple the result out to put it in the standard form of a
polynomical.
Example. The length of a rectangle is 20 inches and increasing at a rate of 3 inches per
minute. The width of a rectangle if 10 inches and decreasing at a rate of 2 inches per
minute. At the moment the area is 200 square inches. How fast is the area changing?
Let
L(t) = length of rectangle at time t
W(t) = width of rectangle at time t
A(t) = L(t)W(t) = area of the rectangle
t = time with now being t = 0.
We are given L(0) = 20, W(0) = 10, L’(0) = 3 and W’(0) = -2. According to the product
rule A'(t) = L(t)W'(t) + W(t)L '(t). Putting in t = 0 one has
A'(0) = L(0)W'(0) + W(0)L '(0) = (20)(-2) + (10)(3) = -10. So the area is
decreasing at a rate of 10 square inches per minute.
Proof of the Product Rule. In order to appreciate the proof of the product rule, it helps
to think in terms of the rectangle example just discussed. So we shall phase the proof in
those terms. One has
dA
L(t+h)W(t+h) - L(t)W(t)
dt = hlim
h
0
Now write
L(t+h) = L(t) + [L(t+h) - L(t)] = L(t) + L
W(t+h) = W(t) + [W(t+h) - W(t)] = W(t) + W
where
L = L(t+h) - L(t)
W = W(t+h) - W(t)
1.5 - 4
Then L(t+h)W(t+h) = [L(t) + L]W(t+h) = L(t)W(t+h) + (L)W(t+h)
= L(t)[W(t) + W] + (L)W(t+h)
= L(t)W(t) + L(t)(W) + (L)W(t+h)
See the picture at the right. So
dA
L(t)(W) + (L)W(t+h)
dt = hlim
h
0
W(t+h) - W(t)
L(t+h) - L(t)
+ lim
lim W(t+h) = L(t)W'(t) + W(t)L '(t)
h
h
h0
h0
h0
= L(t) lim
The reason lim W(t+h) = W(t) is because
h0
W(t+h) - W(t)
= W(t) + 0 W'(t)
h
h0
h0
h0
In other words, it a function has a derivative at a certain value of t, then it is also
continuous at this value of t.
lim W(t+h) = W(t) + lim h
lim
Example. Consider a container with a gas inside. The ideal gas law says
PV = nRT
where
P = pressure of the gas (nt/m2)
V = volume of the container (m3)
weight of gas in gms
n = amount of gas in moles = molecular weight of the gas
R = gas constant = 8.32 nt-m/mole-deg K
T = absolute temperature of the gas (deg K)
Suppose the volume is held constant at 1 m3, so that
P = nRT
Suppose at the moment n = 4 and is increasing at a rate of 2 moles/hr and T = 300 K and
is decreasing at a rate of 3 / hr. What is the pressure and how fast is it changing?
P = (8.32)(4)(300) = 9984 nt/m2. Also, we are given n’ = 2 and T’(0) = -3. According to
the product rule P ' = RnT ' + RTn ' = (8.32)[(4)(-3) + (300)(2)] = (8.32)[-12 + 600]
= (8.32)(588) = 4892.16 nt/m2-hr. So the pressure is increasing at a rate of 4892.16 nt/m2
per hour.
1.5 - 5
Rule #6 (Generalized Product Rule).
d
df
dg
dh
[f(x)g(x)h(x)]
=
g(x)h(x)
+
f(x)
h(x)
+
f(x)g(x)
dx
dx
dx
dx
d
df1
df2
dfn
…
…
…
…
…
dx [f1(x)f2(x) fn(x)] = dx f2(x) fn(x) + f1(x) dx f3(x) fn(x) + + f1(x)f2(x) fn-1(x) dx
In other words, the derivative of a product is a sum of terms where in each term one of
the factors gets differentiated and the others are left alone.
Example.
d
[(x2 + 3)(x3 – 6x)(x4 + 2x2)] = 2x(x3 – 6x)(x4 + 2x2) + (x2 + 3)(3x2 – 6)(x4 + 2x2)
dx
+ (x2 + 3)(x3 – 6x)(4x3 + 4x)
Proof of the generalized product rule in the case n = 3.
d
df
d
dx [f(x)g(x)h(x)] = dx g(x)h(x) + f(x)dx[g(x)h(x)]
df
dg
dh
= dx g(x)h(x) + f(x) [dx h(x) + g(x) dx]
df
dg
dh
= dx g(x)h(x) + f(x) dxh(x) + f(x)g(x) dx
d
df
Rule #7 (The Generalized Power Rule). dx [f(x)]n = n[f(x)]n-1 dx. This holds not only
for integer n, but for any real number n.
Examples.
d 2
4
2
3 d
2
2
3
dx (x –7x + 2) = 4(x –7x + 2) dx (x –7x + 2) = 4(x –7x + 2) (2x – 7)
d
d 2
x
2
1/2
2
-1/2 d
2
2
-1/2
x
+
1
=
(x
+
1)
=
(½)(x
+
1)
(x
+
1)
=
(½)(x
+
1)
(2x)
=
2
dx
dx
dx
x +1
1.5 - 6