Differentiation of Rational FunctionsMathematics 11: Lecture 17

Dan Sloughter

Furman University

October 9, 2007

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 1 / 15

Quotients

I Suppose f and g are differentiable and s(x) =f (x)

g(x).

I Then

s ′(x) = limh→0

s(x + h)− s(x)

h

= limh→0

f (x + h)

g(x + h)− f (x)

g(x)

h

= limh→0

f (x + h)g(x)− f (x)g(x + h)

hg(x)g(x + h)

= limh→0

f (x + h)g(x)− f (x)g(x) + f (x)g(x)− f (x)g(x + h)

hg(x)g(x + h).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 2 / 15

Quotients

I Suppose f and g are differentiable and s(x) =f (x)

g(x).

I Then

s ′(x) = limh→0

s(x + h)− s(x)

h

= limh→0

f (x + h)

g(x + h)− f (x)

g(x)

h

= limh→0

f (x + h)g(x)− f (x)g(x + h)

hg(x)g(x + h)

= limh→0

f (x + h)g(x)− f (x)g(x) + f (x)g(x)− f (x)g(x + h)

hg(x)g(x + h).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 2 / 15

Quotients (cont’d)

I And so

s ′(x) = limh→0

g(x)(f (x + h)− f (x))− f (x)(g(x + h)− g(x))

hg(x)g(x + h)

= limh→0

g(x)f (x + h)− f (x)

h− f (x)

g(x + h)− g(x)

hg(x)g(x + h)

=g(x)f ′(x)− f (x)g ′(x)

(g(x))2.

I Note: we have used the continuity of g to justify

limh→0

g(x + h) = g(x).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 3 / 15

Quotients (cont’d)

I And so

s ′(x) = limh→0

g(x)(f (x + h)− f (x))− f (x)(g(x + h)− g(x))

hg(x)g(x + h)

= limh→0

g(x)f (x + h)− f (x)

h− f (x)

g(x + h)− g(x)

hg(x)g(x + h)

=g(x)f ′(x)− f (x)g ′(x)

(g(x))2.

I Note: we have used the continuity of g to justify

limh→0

g(x + h) = g(x).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 3 / 15

Quotient Rule

I If f and g are differentiable and

s(x) =f (x)

g(x),

then

s ′(x) =g(x)f ′(x)− f (x)g ′(x)

(g(x))2.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 4 / 15

Example

I If

f (x) =4x + 1

5x − 1,

then

f ′(x) =(5x − 1)(4)− (4x + 1)(5)

(5x − 1)2= − 9

(5x − 1)2.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 5 / 15

Example

I If

f (x) =3x2 − 5

x2 + 1,

then

f ′(x) =(x2 + 1)(6x)− (3x2 − 5)(2x)

(x2 + 1)2=

16x

(x2 + 1)2.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 6 / 15

Example

I If

f (x) =1

x2,

then

f ′(x) =(x2)(0)− (1)(2x)

x4= −2x

x4= − 2

x3.

I Note: we could write f (x) = x−2 and f ′(x) = −2x−3.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 7 / 15

Example

I If

f (x) =1

x2,

then

f ′(x) =(x2)(0)− (1)(2x)

x4= −2x

x4= − 2

x3.

I Note: we could write f (x) = x−2 and f ′(x) = −2x−3.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 7 / 15

Powers of x revisited

I Suppose n is a negative integer and let f (x) = xn.

I Then we may also write

f (x) =1

x−n,

and so, using the quotient rule, we have

f ′(x) =(x−n)(0)− (1)(−nx−n−1)

x−2n=

nx−n−1

x−2n= nxn−1.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 8 / 15

Powers of x revisited

I Suppose n is a negative integer and let f (x) = xn.

I Then we may also write

f (x) =1

x−n,

and so, using the quotient rule, we have

f ′(x) =(x−n)(0)− (1)(−nx−n−1)

x−2n=

nx−n−1

x−2n= nxn−1.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 8 / 15

Theorem

I If n is an integer and f (x) = xn, then f ′(x) = nxn−1.

I Example: if

f (x) =5

x7,

then f (x) = 5x−7, so

f ′(x) = −35x−8 = −35

x8.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 9 / 15

Theorem

I If n is an integer and f (x) = xn, then f ′(x) = nxn−1.

I Example: if

f (x) =5

x7,

then f (x) = 5x−7, so

f ′(x) = −35x−8 = −35

x8.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 9 / 15

Example

I If

g(t) =3t2 + 1

t,

then we may write

g(t) = 3t +1

t= 3t + t−1,

and so

g ′(t) = 3− t−2 = 3− 1

t2.

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 10 / 15

Derivative of tangent

I We now have

d

dxtan(x) =

d

dx

(sin(x)

cos(x)

)=

(cos(x))(cos(x))− (sin(x))(− sin(x))

cos2(x)

=cos2(x) + sin2(x)

cos2(x)

=1

cos2(x)

= sec2(x).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 11 / 15

Derivative of cotangent

I And

d

dxcot(x) =

d

dx

(cos(x)

sin(x)

)=

(sin(x))(− sin(x))− (cos(x))(cos(x))

sin2(x)

= −sin2(x) + cos2(x)

sin2(x)

= − 1

sin2(x)

= − csc2(x).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 12 / 15

Derivative of secant

I And

d

dxsec(x) =

d

dx

1

cos(x)

=cos(x)(0)− (− sin(x))

cos2(x)

=sin(x)

cos2(x)

=1

cos(x)

sin(x)

cos(x)

= sec(x) tan(x).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 13 / 15

Derivative of cosecant

I And, finally,

d

dxcsc(x) =

d

dx

1

sin(x)

=sin(x)(0)− cos(x)

sin2(x)

= − cos(x)

sin2(x)

= − 1

sin(x)

cos(x)

sin(x)

= − csc(x) cot(x).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 14 / 15

Examples

I If f (x) = sec(x) tan(x), then

f ′(x) = sec(x) sec2(x) + tan(x) sec(x) tan(x)

= sec3(x) + sec(x) tan2(x).

I If g(t) =cot(t)

t2, then

g ′(t) =−t2 csc2(t)− 2t cot(t)

t4= − 1

t2csc2(t)− 2

t3cot(t).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 15 / 15

Examples

I If f (x) = sec(x) tan(x), then

f ′(x) = sec(x) sec2(x) + tan(x) sec(x) tan(x)

= sec3(x) + sec(x) tan2(x).

I If g(t) =cot(t)

t2, then

g ′(t) =−t2 csc2(t)− 2t cot(t)

t4= − 1

t2csc2(t)− 2

t3cot(t).

Dan Sloughter (Furman University) Differentiation of Rational Functions October 9, 2007 15 / 15