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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