2.6 more computations of derivatives

More Computations of Derivatives

We list the derivative rules we have so far.

We assume that the derivatives exist in all the
theorems below and the prime ( )' operation is
differentiation with respect to x.

The ± and Constant–Multiple Derivative Rules

Let f(x) and g(x) be two functions then
i. (f(x)±g(x)) ' = f '(x)±g '(x)
ii. (cf(x)) ' = c*f '(x) where c is a constant.

i. (f(x)±g(x)) ' = f '(x)±g '(x)
The Product and Quotient Rules of Derivatives

i. (f(x)±g(x)) ' = f '(x)±g '(x)
The Product and Quotient Rules of Derivatives
Write f for f(x) and g for g(x) then
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
The Product Rule
The Quotient Rule

Derivatives of Constants and Power Functions
i. If f(x) = c, a constant function, then f '(x) = (c)' = 0.
ii. If f(x) = xP, then f '(x) = PxP–1 for a constant P ≠ 0.

From the above rules and the derivatives of the
monomials we are able to easily calculate the
derivative of polynomials and rational functions.

But in order to take the derivatives of all elementary
functions, we also need to know the derivatives of the
other basic functions i.e. functions of the trig. family,
the log and exponential functions.

But in order to take the derivatives of all elementary
functions, we also need to know the derivatives of the
other basic functions i.e. functions of the trig. family,
the log and exponential functions.
More importantly we need to know how the derivative
of the composition of two functions such as sin(x2) may
be expressed as the derivative of each component.

Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)

i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
d cos(x)
dx = –sin(x)
In the following verification of i, we write s(x) for sin(x)
and c(x) for cos(x).

i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
d cos(x)
dx = –sin(x)
and c(x) for cos(x). We need the following limit–results.

i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
d cos(x)
dx = –sin(x)
lim
sin(h)
h
= 1
h→0
lim
cos(h) – 1
h
= 0
h→0
(HW)

i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
d cos(x)
dx = –sin(x)
lim
sin(h)
h
= 1
h→0
cos(h) – 1
Expand the difference quotient of sin(x) by the Sum of
Angles Formula for sine.
s(x + h) – s(x)
h =
lim
h
= 0
h→0
(HW)

i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
d cos(x)
dx = –sin(x)
lim
sin(h)
h
= 1
h→0
cos(h) – 1
s(x + h) – s(x)
h =
s(x)c(h) + c(x)s(h) – s(x)
h
lim
h
= 0
h→0
(HW)

i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
d cos(x)
dx = –sin(x)
lim
sin(h)
h
= 1
h→0
cos(h) – 1
s(x + h) – s(x)
h =
s(x)c(h) + c(x)s(h) – s(x)
h
= s(x)c(h) – s(x) + c(x)s(h)
h
lim
h
= 0
h→0
(HW)

i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
d cos(x)
dx = –sin(x)
lim
sin(h)
h
= 1
h→0
cos(h) – 1
s(x + h) – s(x)
h =
s(x)c(h) + c(x)s(h) – s(x)
h
= s(x)c(h) – s(x) + c(x)s(h)
h
= s(x)(c(h) – 1) + c(x)s(h)
h
lim
h
= 0
h→0
(HW)

Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h

Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
c(x)s(h)
+ lim
h→0
h

Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
0 1
c(x)s(h)
+ lim
h→0
h

Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
0 1
c(x)s(h)
+ lim
h→0
h = cos(x)

Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
0 1
c(x)s(h)
+ lim
h→0
h = cos(x)
Therefore d sin(x)
dx = cos(x)

Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
0 1
c(x)s(h)
+ lim
h→0
h = cos(x)
Therefore d sin(x)
dx = cos(x)
The verification of ii is similar and we leave it as an
exercise.

Derivatives of Other Trig. Functions
The derivatives of the other trig. functions may be
obtained by the quotient rule and the derivatives of
sine and cosine.
tan'(x) = sec2(x)
d tan(x)
dx = sec2(x)
cot'(x) = –csc2(x)
d cot(x)
dx = –csc2(x)
sec'(x) = sec(x)tan(x)
csc'(x) = –csc(x)cot(x)
d sec(x)
dx = sec(x)tan(x)
d csc(x)
dx = –csc(x)cot(x)

Example A. Find the derivatives of the following
functions.
a.
d tan(x)
dx
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)

functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)

functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
=
c(x)s'(x) – s(x)c'(x)
(xN)' = NxN–1
c2(x) sin'(x) = cos(x)
cos'(x) = –sin(x)

functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
=
c(x)s'(x) – s(x)c'(x)
(xN)' = NxN–1
cos'(x) = –sin(x)
c(x) –s(x)

functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
=
c(x)s'(x) – s(x)c'(x)
(xN)' = NxN–1
cos'(x) = –sin(x)
c(x) –s(x)
=
c2(x) + s2(x)
c2(x)

functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
=
c(x)s'(x) – s(x)c'(x)
(xN)' = NxN–1
cos'(x) = –sin(x)
c(x) –s(x)
=
c2(x) + s2(x)
c2(x)
=
1
c2(x)
= sec2(x)

b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)

b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'

b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'
s2(x)
=
0 c(x)
s(x)(1)' – 1*s'(x)

b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'
s2(x)
=
0 c(x)
s(x)(1)' – 1*s'(x)
=
–c(x)
s2(x)
= –csc(x)cot(x)

b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'
s2(x)
=
0 c(x)
s(x)(1)' – 1*s'(x)
=
–c(x)
s2(x)
= –csc(x)cot(x)
Replacing s(x) by a generic function f(x) in the above
proof we get the a short cut for the derivative of 1/f.

b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'
s2(x)
=
0 c(x)
s(x)(1)' – 1*s'(x)
=
–c(x)
s2(x)
= –csc(x)cot(x)
Replacing s(x) by a generic function f(x) in the above
proof we get the a short cut for the derivative of 1/f.
–f '
(Derivative of 1/f) (1/f)' =
f2
d(1/f)
dx
–df/dx
= f2

Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)

(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'

(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= tan'(x)*tan(x) + tan(x)*tan'(x)

(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= 2tan(x)[tan'(x)]

(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= 2tan(x)[tan'(x)]
= 2tan(x)sec2(x)

(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= 2tan(x)[tan'(x)]
= 2tan(x)sec2(x)
Again replacing tan(x) by a function f(x) in the above
proof we get the a short cut for the derivative f2(x).

(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= 2tan(x)[tan'(x)]
(Derivative of f2(x)) (f2(x))' = 2f(x)f '(x)
d(f2)= 2f
dx
df
dx
= 2tan(x)sec2(x)

(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= 2tan(x)[tan'(x)]
= 2tan(x)sec2(x)
(Derivative of f2(x)) (f2(x))' = 2f(x)f '(x)
d(f2)= 2f
dx
df
dx
Note the attached tail end derivative factor f '(x).

The Chain Rule is the rule for calculating the
derivative of the composition of two functions.

It demonstrates how the derivative of the composition
can be assembled from the derivatives of the
component functions.

(The Chain Rule) Let u = u(v) and v = v(x),
the derivative of the composition function
(u ○ v)(x) = u( v(x) ) is the product of the derivatives
of u and v,

of u and v, that is du
dv
dv
dx
du =
dx

of u and v, that is du
dv
dv
dx
du =
dx
Let’s use the last example again.
Let u = u(v) = v2,
and v = v(x) = tan(x),
the composition is u(v(x)) = tan2(x).

We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)

We have that
u(v(x)) = tan2(x)
d tan2(x)
dx
=
Hence
du
dv
dv
dx

We have that
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
Hence
du
dv
dv
dx

We have that
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
d tan(x)
dx
Hence
du
dv
dv
dx

We have that
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
d tan(x)
dx
Hence
du
dv
dv
dx
the back–
derivative

We have that
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
d tan(x)
dx
Hence
du
dv
dv
dx
the back–
derivative
= (2v) sec2(x)

We have that
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
d tan(x)
dx
Hence
du
dv
dv
dx
the back–
derivative
= (2v) sec2(x)
= 2tan(x) sec2(x)

We have that
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
=
d tan(x)
dx
Hence
du
dv
dv
dx
= (2v) sec2(x)
the back–
derivative
= 2tan(x) sec2(x)
The name “Chain Rule” came from the fact that the
rule is often applied many times with the
back–derivative spinning off another back–derivative
which spins off another etc…

Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
b. Let y = sin3(x) find dy/dx.

Example C.
We have that y = sin(u) = sin(x3).

Example C.
dy
dx =
dy
du
du
dx
Hence

Example C.
dy
dx =
dy
du
du
dx =
dsin(u)
du
Hence

Example C.
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
Hence
the back–
derivative

Example C.
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
Hence
the back–
derivative

Example C.
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
Hence
We have that y = u3 where u = sin(x).
the back–
derivative

Example C.
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
dy
dx =
dy
du
du
dx
Hence
Hence
the back–
derivative

Example C.
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
dy
dx =
dy
du
du
dx =
du3
du
d sin(x)
dx
Hence
Hence
the back–
derivative
the back–
derivative

Example C.
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
dy
dx =
dy
du
du
dx =
du3
du
d sin(x)
dx
= 3u2 cos(x) = 3sin2(x) cos(x)
Hence
Hence
the back–
derivative
the back–
derivative

(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dy
du
=
..
dv
dx du
dw
dx

dy
dy
du
=
..
dv
dx du
dw
dx
c. Find the derivative of cos[(x2+1)3]

dy
dy
du
=
..
dv
dx du
dw
dx
We have that y = cos(u),

dy
dy
du
=
..
dv
dx du
dw
dx
where u = v3,

dy
dy
du
=
..
dv
dx du
dw
dx
where u = v3,
where v = (x2 + 1).

dy
dy
du
=
..
dv
dx du
dw
dx
where u = v3,
where v = (x2 + 1).
dy
dx =
dy
du
du
dv
dv
Hence dx

dy
dx =
where u = v3,
where v = (x2 + 1).
dy
dx =
dy
du
du
dv
dv
Hence dx
= –sin(u) (3v2) (2x)
dy
du
du
dw
dv
dx
..

dy
dx =
where u = v3,
where v = (x2 + 1).
dy
dx =
dy
du
du
dv
dv
Hence dx
= –sin(u) (3v2) (2x)
= –sin(v3) (3v2) (2x)
dy
du
du
dw
dv
dx
..

dy
dx =
where u = v3,
where v = (x2 + 1).
dy
dx =
dy
du
du
dv
dv
Hence dx
= –sin(u) (3v2) (2x)
= –sin(v3) (3v2) (2x)
= –sin[(x2 + 1)3] [3(x2 + 1)2](2x)
dy
du
du
dw
dv
dx
..

The prime notation of the Chain Rule may look
confusing.
(u ○ v)'(x) = u'(v(x))v'(x)

confusing.
derivative taken
with respect to x
derivative taken
with respect to x
derivative taken
with respect to v
(u ○ v)'(x) = u'(v(x))v'(x)

confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
is the derivative of u with
dv v=v(x)
respect to v, evaluated at v (x).
derivative taken
with respect to x
derivative taken
with respect to v

confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
dv v=v(x)
derivative taken
with respect to x
derivative taken
with respect to v
Example D. Let u(v) = cos(v) and v(x) = x2, find the
slope of the tangent at x = √π for (u ○ v)(x).

confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
dv v=v(x)
derivative taken
with respect to x
derivative taken
with respect to v
(u ○ v)'(x) = u'(v(x))v'(x) =

confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
dv v=v(x)
derivative taken
with respect to x
derivative taken
with respect to v
(u ○ v)'(x) = u'(v(x))v'(x) = –sin(v) 2x,

confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
dv v=v(x)
derivative taken
with respect to x
derivative taken
with respect to v
(u ○ v)'(x) = u'(v(x))v'(x) = –sin(v) 2x, at x = √π,
(u ○ v)'(√π) = –sin(v) 2x x = √π,

confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
dv v=v(x)
derivative taken
with respect to x
derivative taken
with respect to v
(u ○ v)'(x) = u'(v(x))v'(x) = –sin(v) 2x, at x = √π,
(u ○ v)'(√π) = –sin(v) 2x x = √π,
= –sin(√π2) 2√π = 0

The Chain Rule gives the multiplying effect of the
rates of change under the composition operation.

Suppose y = g(u) = 3u and u = f(x) = 5x.

The slope of y = g(u) = 3u as a function in u is 3.
The slope u = 5x with respect to x is 5.

The slope of y = g(u) = 3u as a function in u is 3.
The slope u = 5x with respect to x is 5.
The composition y = g(f(x)) as a function in x is
y = 3(5x) = 15x which has slope to x as (3)(5) = 15.

2.6 more computations of derivatives

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