Section 2.3 Basic Differentiation Formulas 2010 Kiryl Tsishchanka

Basic Differentiation Formulas

DERIVATIVE OF A CONSTANT FUNCTION:

d

dx(c) = 0 or c′ = 0

Proof: Suppose f(x) = c, then

f ′(x) = limh→0

f(x+ h)− f(x)

h= lim

h→0

c− c

h= lim

h→0

0

h= lim

h→00 = 0

EXAMPLES:

1′ = 0, 5′ = 0, 0′ = 0, (−7/9)′ = 0, π′ = 0,

(

1 +√5

2

)

′

= 0, (x 3√x+ 1− x4/3)′ = 0

THE POWER RULE: If n is a positive integer, then

d

dx(xn) = nxn−1 or (xn)′ = nxn−1

Proof: Before we prove this result rigorously, let us consider cases n = 2, 3, 4.

If n = 2, then

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→a

x2 − a2

x− a= lim

x→a

(x− a)(x+ a)

x− a= lim

x→a(x+ a) = a+ a = 2a

If n = 3, then

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→a

x3 − a3

x− a= lim

x→a

(x− a)(x2 + xa+ a2)

x− a

= limx→a

(x2 + xa+ a2) = a2 + a · a+ a2 = 3a2

If n = 4, then

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→a

x4 − a4

x− a= lim

x→a

(x− a)(x3 + x2a+ xa2 + a3)

x− a

= limx→a

(x3 + x2a+ xa2 + a3) = a3 + a2 · a+ a · a2 + a3 = 4a3

In general, we have

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→a

xn − an

x− a= lim

x→a

(x− a)(xn−1 + xn−2a+ . . .+ xan−2 + an−1)

x− a

= limx→a

(xn−1 + xn−2a+ . . .+ xan−2 + an−1) = nan−1

1

Section 2.3 Basic Differentiation Formulas 2010 Kiryl Tsishchanka

EXAMPLES:

(a) If f(x) = x2, then f ′(x) = (x2)′ = [n = 2] = 2x2−1 = 2x1 = 2x.

(b) If f(x) = x9, then f ′(x) = (x9)′ = [n = 9] = 9x9−1 = 9x8.

(c) If f(x) = x, then f ′(x) = (x1)′ = [n = 1] = 1 · x1−1 = 1 · x0 = 1 · 1 = 1.

THE POWER RULE (GENERAL VERSION): If n is any real number, then

d

dx(xn) = nxn−1 or (xn)′ = nxn−1

EXAMPLES:

(a) If f(x) = x−4, then f ′(x) = (x−4)′ = [n = −4] = (−4)x−4−1 = −4x−5.

(b) If f(x) =1

x, then f ′(x) = (x−1)′ = [n = −1] = (−1)x−1−1 = −x−2 = −

1

x2.

(c) If f(x) =√x, then f ′(x) = (x1/2)′ = [n = 1/2] =

1

2x1/2−1 =

1

2x−1/2 =

1

2√x.

(d) If f(x) = x2 3√x, then

f ′(x) = (x2 · x1/3)′ = (x2+1/3)′ = (x7/3)′ = [n = 7/3] =7

3x7/3−1

=7

3x4/3 =

{

73√x4

3=

73√x3+1

3=

73√x3 · x3

=7

3√x3 · 3

√x

3

}

=7x 3

√x

3

or

=

{

7x4

3

3=

7x3+1

3

3=

7x3

3+ 1

3

3=

7x3

3 · x 1

3

3

}

=7x 3

√x

3

(e) If f(x) =1

3√x2

, then

f ′(x) =

(

1

x2/3

)

′

= (x−2/3)′ = [n = −2/3] = −2

3x−2/3−1 = −

2

3x−5/3

=

{

−2

3x5/3= −

2

33√x5

= −2

33√x3+2

= −2

33√x3 · x2

= −2

33√x3 · 3

√x2

}

= −2

3x3√x2

or

=

{

−2

3x5

3

= −2

3x3+2

3

= −2

3x3

3+ 2

3

= −2

3x3

3 · x 2

3

}

= −2

3x3√x2

(f) Find f ′(x) if f(x) =4√x

x−1√x5

.

2

Section 2.3 Basic Differentiation Formulas 2010 Kiryl Tsishchanka

(f) If f(x) =4√x

x−1√x5

, then

f ′(x) =

(

x1/4

x−1 · x5/2

)′

=

(

x1/4

x−1+5/2

)′

=(

x1/4−(−1+5/2))′

=(

x1/4+1−5/2)′

=(

x−5/4)′

= [n = −5/4]

= −5

4x−5/4−1 = −

5

4x−9/4 =

{

−5

44√x9

= −5

44√x8+1

= −5

44√x8 · x

= −5

44√x8 · 4

√x

}

= −5

4x2 4√x

or

=

{

−5

4x9

4

= −5

4x8+1

4

= −5

4x8

4+ 1

4

= −5

4x8

4x1

4

}

= −5

4x2 4√x

THE CONSTANT MULTIPLE RULE: If c is a constant and f is a differentiable function, then

d

dx[cf(x)] = c

d

dxf(x) or (cf(x))′ = cf ′(x) or (cf)′ = cf ′

Proof: We have

(cf(x))′ = limh→0

cf(x+ h)− cf(x)

h= lim

h→0

c[f(x+ h)− f(x)]

h= c · lim

h→0

f(x+ h)− f(x)

h= cf ′(x)

EXAMPLE: If f(x) =2

3 5√x, then

f ′(x) =

(

2

3x−1/5

)

′

=2

3

(

x−1/5)′

=2

3

(

−1

5

)

x−1/5−1 = −2

15x−6/5

EXAMPLE: Find equations of the tangent line and normal line to the curve y = 2x3 ·√x at

the point (1, 2).

Solution: A point-slope equation of a line is

y − y0 = m(x− x0)

where m is the slope. Since x0 and y0 are given (x0 = 1 and y0 = 2), we only have to find theslope. We have

f ′(x) = (2x3 · x1/2)′ = (2x3+1/2)′ = 2(x7/2)′ = 2 ·7

2x7/2−1 = 7x5/2

therefore the slope of the tangent line at (1, 2) is f ′(1) = 7 · 15/2 = 7. So, an equation of thetangent line is

y − 2 = 7(x− 1)

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal

of 7, that is, −1

7. From this it follows that an equation of the normal line is

y − 2 = −1

7(x− 1)

3

Section 2.3 Basic Differentiation Formulas 2010 Kiryl Tsishchanka

THE SUM/DIFFERENCE RULE: If f and g are both differentiable functions, then

d

dx[f(x)± g(x)] =

d

dxf(x)±

d

dxg(x)

or(f(x)± g(x))′ = f ′(x)± g′(x) or (f ± g)′ = f ′ ± g′

Proof: We have

(f(x)± g(x))′ = limh→0

(f(x+ h)± g(x+ h))− (f(x)± g(x))

h

= limh→0

[f(x+ h)− f(x)]± [g(x+ h)− g(x)]

h

= limh→0

f(x+ h)− f(x)

h± lim

h→0

g(x+ h)− g(x)

h

= f ′(x)± g′(x)

EXAMPLE: If f(x) =3x2 − 5

√x

6x4, then

f ′(x) =

(

3x2 − 5x1/2

6x4

)′

=

(

3x2

6x4−

5x1/2

6x4

)′

=

(

3

6x2−4 −

5

6x1/2−4

)

′

=

(

1

2x−2 −

5

6x−7/2

)

′

=

(

1

2x−2

)

′

−(

5

6x−7/2

)

′

=1

2

(

x−2)

′ −5

6

(

x−7/2)′

=1

2· (−2)x−2−1 −

5

6·(

−7

2

)

· x−7/2−1 = −x−3 +35

12x−9/2

REMARK: We can combine two previous rules in one. If c1, c2 are constants and f, g are bothdifferentiable functions, then

d

dx[c1f(x)± c2g(x)] = c1

d

dxf(x)± c2

d

dxg(x)

or(c1f(x)± c2g(x))

′ = c1f′(x)± c2g

′(x) or (c1f ± c2g)′ = c1f

′ ± c2g′

EXAMPLE: If f(x) =3x2 − 5

√x

6x4, then

f ′(x) =

(

3x2 − 5x1/2

6x4

)′

=

(

3x2

6x4−

5x1/2

6x4

)′

=

(

3

6x2−4 −

5

6x1/2−4

)

′

=

(

1

2x−2 −

5

6x−7/2

)

′

=1

2

(

x−2)

′ −5

6

(

x−7/2)′

=1

2· (−2)x−2−1 −

5

6·(

−7

2

)

· x−7/2−1 = −x−3 +35

12x−9/2

4

Section 2.3 Basic Differentiation Formulas 2010 Kiryl Tsishchanka

THE DERIVATIVE OF THE SINE AND COSINE FUNCTIONS: We have

d

dx(sin x) = cos x or (sin x)′ = cos x

andd

dx(cos x) = − sin x or (cosx)′ = − sin x

Proof: We have

(sin x)′ = limh→0

sin(x+ h)− sin x

h

[We use sin(α + β) = sinα cos β + cosα sin β]

= limh→0

sin x cosh+ cos x sinh− sin x

h= lim

h→0

(

sin x cosh− sin x

h+

cos x sinh

h

)

= limh→0

(

sin x(cosh− 1)

h+ cos x ·

sinh

h

)

= sin x limh→0

cosh− 1

h+ cos x lim

h→0

sinh

h

=sin x · 0 + cos x · 1 = cos x

In the same way we prove that (cos x)′ = − sin x.

EXAMPLE: If f(x) = 3 sin x− 4 cos x, then

f ′(x) = (3 sin x− 4 cos x)′ = 3(sin x)′ − 4(cos x)′ = 3 cos x− 4(− sin x) = 3 cos x+ 4 sin x

EXAMPLE: If f(x) = sin x, then

f ′(x) = cos x

f ′′(x) = (cosx)′ = − sin x

f ′′′(x) = (− sin x)′ = − cos x

f ′′′′(x) = (− cos x)′ = −(− sin x) = sin x

Thereforef(x) = f (4)(x) = f (8)(x) = f (12)(x) = f (16)(x) = . . . = sin x

f ′(x) = f (5)(x) = f (9)(x) = f (13)(x) = f (17)(x) = . . . = cos x

f ′′(x) = f (6)(x) = f (10)(x) = f (14)(x) = f (18)(x) = . . . = − sin x

f ′′′(x) = f (7)(x) = f (11)(x) = f (15)(x) = f (19)(x) = . . . = − cos x

For instance, it immediately follows from here that

f (2010)(x) = f (2008+2)(x) = f (4·502+2) = f ′′(x) = − sin x

5

Section 2.3 Basic Differentiation Formulas 2010 Kiryl Tsishchanka

Appendix

limh→0

cosh− 1

h=

[

0

0

]

A= lim

h→0

(cosh− 1)(cosh+ 1)

h(cosh+ 1)A= lim

h→0

cos2 h− 1

h(cosh+ 1)

A= lim

h→0

−(1− cos2 h)

h(cosh+ 1)

T= lim

h→0

− sin2 h

h(cosh+ 1)

A= lim

h→0

(

sinh

h·

− sinh

cosh+ 1

)

C= lim

h→0

sinh

h· limh→0

− sinh

cosh+ 1

C= 1 · lim

h→0

− sinh

cosh+ 1

DSP=

− sin 0

1 + cos 0

=0

1 + 1

= 0

6