mat01a1: derivatives of log functions and logarithmic … · mat01a1: derivatives of log functions...

MAT01A1: Derivatives of Log Functions andLogarithmic Differentiation

Dr Craig

Week: 4 May 2020

Reminder: the Chain Rule

If f and g are both differentiable and

F = f ◦ g is the composite function

defined by

F (x) = f (g(x)),

then F is differentiable and F ′ is given by

F ′(x) = f ′(g(x)).g′(x)

Two other reminders

I ddx (a

x) = ax ln a

I When we use implicit differentiation, we

regard y as a function of x.

In the slides that follow, I will sometimes use

y = lnx and sometimes y = `n x to denote

y = loge x. It is often good to use `n x when

writing by hand so you don’t confuse the

function with other similar symbols.

Derivatives of Log Functions

We can use implicit differentiation to find

the derivative of the log function y = loga x.

Proof: Let y = loga x.

∴ ay = x

dx(ay) =

∴ ay · ln a · y′ = 1

∴ y′ =1

ay · ln a

∴ y′ =1

x ln a

∴ ay = x

dx(ay) =

∴ ay · ln a · y′ = 1

∴ y′ =1

ay · ln a

∴ y′ =1

x ln a

∴ ay = x

dx(ay) =

∴ ay · ln a · y′ = 1

∴ y′ =1

ay · ln a

∴ y′ =1

x ln a

∴ ay = x

dx(ay) =

∴ ay · ln a · y′ = 1

∴ y′ =1

ay · ln a

∴ y′ =1

x ln a

∴ ay = x

dx(ay) =

∴ ay · ln a · y′ = 1

∴ y′ =1

ay · ln a

∴ y′ =1

x ln a

∴ ay = x

dx(ay) =

∴ ay · ln a · y′ = 1

∴ y′ =1

ay · ln a

∴ y′ =1

x ln a

dx(loga x) =

x ln a

Notice that if a = e then we have

dx(lnx) =

An example (with chain rule):

Differentiate: y = log10(2 + cosx)

dx(loga x) =

x ln a

dx(lnx) =

dx(loga x) =

x ln a

dx(lnx) =

Example: Differentiate y = log10(2 + cosx).

Solution:

dx(log10(2 + cosx))

∴ y′ =1

(2 + cosx) ln 10· ddx

(2 + cosx)

∴ y′ =− sinx

(2 + cosx) ln 10

Solution:

dx(log10(2 + cosx))

∴ y′ =1

(2 + cosx)

∴ y′ =− sinx

(2 + cosx) ln 10

Solution:

dx(log10(2 + cosx))

∴ y′ =1

(2 + cosx)

∴ y′ =− sinx

(2 + cosx) ln 10

Solution:

dx(log10(2 + cosx))

∴ y′ =1

(2 + cosx)

∴ y′ =− sinx

(2 + cosx) ln 10

Derivative of y = lnx

dx(loga x) =

x ln a⇒ d

dx(lnx) =

Example: Differentiate y = ln(x3 + 1).

dx(ln(x3+1)) =

x3 + 1· ddx

(x3+1) =3x2

x3 + 1

In general, if we combine derivatives of log

functions with the chain rule, we get

dx(ln(g(x))) =

g′(x)

dx(loga x) =

x ln a⇒ d

dx(lnx) =

dx(ln(x3+1)) =

x3 + 1· ddx

(x3+1) =3x2

x3 + 1

dx(ln(g(x))) =

g′(x)

dx(loga x) =

x ln a⇒ d

dx(lnx) =

dx(ln(x3+1)) =

x3 + 1· ddx

(x3+1) =3x2

x3 + 1

dx(ln(g(x))) =

g′(x)

dx(loga x) =

x ln a⇒ d

dx(lnx) =

dx(ln(x3+1)) =

x3 + 1· ddx

(x3+1) =3x2

x3 + 1

dx(ln(g(x))) =

g′(x)

More examples:

Differentiate the following:

1. y = ln(sinx)

2. f (x) =√lnx

3. g(x) = ln

(x + 1√x− 2

Solutions:

dx(ln(sinx)) =

sinx· ddx

(sinx) = cotx

More examples:

1. y = ln(sinx)

2. f (x) =√lnx

3. g(x) = ln

(x + 1√x− 2

)Solutions:

dx(ln(sinx)) =

sinx· ddx

(sinx) = cotx

More examples:

1. y = ln(sinx)

2. f (x) =√lnx

3. g(x) = ln

(x + 1√x− 2

)Solutions:

dx(ln(sinx)) =

sinx· ddx

(sinx) = cotx

Solutions:

2. f ′(x) =1

2√lnx· ddx

(lnx) =1

2x√lnx

x + 1√x− 2

(ln(x + 1)− ln

√x− 2

dxln(x + 1)− 1

dxln(x− 2)

x + 1− 1

2(x− 2)=

x− 5

2(x + 1)(x− 2)

Solutions:

2. f ′(x) =1

2√lnx· ddx

(lnx) =1

2x√lnx

x + 1√x− 2

(ln(x + 1)− ln

√x− 2

dxln(x + 1)− 1

dxln(x− 2)

x + 1− 1

2(x− 2)=

x− 5

2(x + 1)(x− 2)

Solutions:

2. f ′(x) =1

2√lnx· ddx

(lnx) =1

2x√lnx

x + 1√x− 2

(ln(x + 1)− ln

√x− 2

dxln(x + 1)− 1

dxln(x− 2)

x + 1− 1

2(x− 2)=

x− 5

2(x + 1)(x− 2)

Solutions:

2. f ′(x) =1

2√lnx· ddx

(lnx) =1

2x√lnx

x + 1√x− 2

(ln(x + 1)− ln

√x− 2

dxln(x + 1)− 1

dxln(x− 2)

x + 1− 1

2(x− 2)=

x− 5

2(x + 1)(x− 2)

Solutions:

2. f ′(x) =1

2√lnx· ddx

(lnx) =1

2x√lnx

x + 1√x− 2

(ln(x + 1)− ln

√x− 2

dxln(x + 1)− 1

dxln(x− 2)

x + 1− 1

2(x− 2)

=x− 5

2(x + 1)(x− 2)

Solutions:

2. f ′(x) =1

2√lnx· ddx

(lnx) =1

2x√lnx

x + 1√x− 2

(ln(x + 1)− ln

√x− 2

dxln(x + 1)− 1

dxln(x− 2)

x + 1− 1

2(x− 2)=

x− 5

2(x + 1)(x− 2)

A graph of the last example:

x+ 1√x− 2

x− 5

2(x+ 1)(x− 2)(x > 2)

Differentiation so far:

I Product rule: (f.g)′ = f ′g + g′f

I Quotient rule:

f ′g − g′f

I Trig derivatives (from special limits)

I Chain Rule: F = f ◦ g,

F ′(x) = f ′(g(x)).g′(x)

I Implicit differentiation: treat y as a

function of x

I Inverse Trig derivatives

I Derivatives of Log functions

The function f (x) = `n(x) is only defined

for x > 0. Hence its derivative f ′(x) =1

only defined for x > 0. What about the

function f (x) = `n|x|?

f (x) = `n|x| f ′(x) =1

x(both defined for x ∈ (−∞, 0) ∪ (0,∞))

What do we know so far about ddx(a

Let y = ab.

I If a, b ∈ R, then y′ = 0.

I If a = f (x), b ∈ R, then we apply the

chain rule with the power rule to get

y′ =d

(f (x)b

)= b[f (x)]b−1f ′(x)

I If a ∈ R, b = g(x), we apply the chain

rule and ddx(a

x) to get

y′ =d

(ag(x)

)= ag(x)(ln a)g′(x)

Let y = ab.

y′ =d

(f (x)b

)= b[f (x)]b−1f ′(x)

rule and ddx(a

x) to get

y′ =d

(ag(x)

Let y = ab.

y′ =d

(f (x)b

)= b[f (x)]b−1f ′(x)

rule and ddx(a

x) to get

y′ =d

(ag(x)

Let y = ab.

y′ =d

(f (x)b

)= b[f (x)]b−1f ′(x)

rule and ddx(a

x) to get

y′ =d

(ag(x)

Logarithmic Differentiation:

Consider the function

y = x2x+3

How do we differentiate this? Do we use the

power rule? Or the rule for exponential

functions?

Neither!

y = x2x+3

How do we differentiate this?

Do we use the

functions?

Neither!

y = x2x+3

power rule?

Or the rule for exponential

functions?

Neither!

y = x2x+3

functions?

Neither!

y = x2x+3

functions?

Neither!

Steps in logarithmic differentiation:

1. Take natural logarithms of both sides of

y = f (x) and use the Log Laws to

simplify the result.

2. Differentiate implicitly with respect to x.

3. Solve for y′.

Example: Differentiate y = (cosx)x

∴ `n(y) = `n ((cosx)x)

= x · `n(cosx)

Now we differentiate both sides:

dx(`n(y)) =

dx(x)`n(cosx) + x

dx(`n(cosx))

∴y′

y= `n(cosx) + x · − sinx

∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)

∴ `n(y) = `n ((cosx)x)

= x · `n(cosx)

dx(`n(y)) =

dx(x)`n(cosx) + x

dx(`n(cosx))

∴y′

∴ `n(y) = `n ((cosx)x)

= x · `n(cosx)

dx(`n(y)) =

dx(x)`n(cosx) + x

dx(`n(cosx))

∴y′

∴ `n(y) = `n ((cosx)x)

= x · `n(cosx)

dx(`n(y)) =

dx(x)`n(cosx) + x

dx(`n(cosx))

∴y′

∴ `n(y) = `n ((cosx)x)

= x · `n(cosx)

dx(`n(y)) =

dx(x)`n(cosx) + x

dx(`n(cosx))

∴y′

∴ `n(y) = `n ((cosx)x)

= x · `n(cosx)

dx(`n(y)) =

dx(x)`n(cosx) + x

dx(`n(cosx))

∴y′

∴ `n(y) = `n ((cosx)x)

= x · `n(cosx)

dx(`n(y)) =

dx(x)`n(cosx) + x

dx(`n(cosx))

∴y′

∴ y′ = y (`n(cosx)− x · tanx)

∴ y′ = (cosx)x (`n(cosx)− x · tanx)

∴ `n(y) = `n ((cosx)x)

= x · `n(cosx)

dx(`n(y)) =

dx(x)`n(cosx) + x

dx(`n(cosx))

∴y′

More examples:

y = (sinx)x3

and y = x√x

Try to find y′ on your own first. The

solutions are on the next two slides.

Example: Differentiate y = (sinx)x3.

Solution:

ln(y)=ln((sinx)x

3)=x3 ln(sinx)

Differentiating both sides gives:

dx(ln(y)) =

dx(x3) ln(sinx) + x3 · d

dx(ln(sinx))

∴y′

y= 3x2 ln(sinx) + x3

sinx· ddx

(sinx)

∴ y′ = y(3x2 ln(sinx) + x3 cotx

)∴ y′ = (sinx)x

3(3x2 ln(sinx) + x3 cotx

Solution: ln(y)=ln((sinx)x

3)=x3 ln(sinx)

dx(ln(y)) =

dx(ln(sinx))

∴y′

sinx· ddx

(sinx)

)∴ y′ = (sinx)x

3)=x3 ln(sinx)

dx(ln(y)) =

dx(ln(sinx))

∴y′

sinx· ddx

(sinx)

)∴ y′ = (sinx)x

3)=x3 ln(sinx)

dx(ln(y)) =

dx(ln(sinx))

∴y′

sinx· ddx

(sinx)

)∴ y′ = (sinx)x

3)=x3 ln(sinx)

dx(ln(y)) =

dx(ln(sinx))

∴y′

sinx· ddx

(sinx)

)∴ y′ = (sinx)x

3)=x3 ln(sinx)

dx(ln(y)) =

dx(ln(sinx))

∴y′

sinx· ddx

(sinx)

∴ y′ = (sinx)x3(3x2 ln(sinx) + x3 cotx

3)=x3 ln(sinx)

dx(ln(y)) =

dx(ln(sinx))

∴y′

sinx· ddx

(sinx)

)∴ y′ = (sinx)x

Example: Differentiate y = x√x.

Solution:

`n(y) = `n(x√x)=√x `n(x)

dx(`n(y)) =

dx(√x)`n(x) +

√x · d

dx(`n(x))

∴y′

2√x`n(x) +

√x · 1

∴y′

+1√x

∴ y′ =(x√x)(`n(x) + 2√

Solution: `n(y) = `n(x√x)=√x `n(x)

dx(`n(y)) =

dx(√x)`n(x) +

√x · d

dx(`n(x))

∴y′

2√x`n(x) +

√x · 1

∴y′

+1√x

∴ y′ =(x√x)(`n(x) + 2√

dx(`n(y)) =

dx(√x)`n(x) +

√x · d

dx(`n(x))

∴y′

2√x`n(x) +

√x · 1

∴y′

+1√x

∴ y′ =(x√x)(`n(x) + 2√

dx(`n(y)) =

dx(√x)`n(x) +

√x · d

dx(`n(x))

∴y′

2√x`n(x) +

√x · 1

∴y′

+1√x

∴ y′ =(x√x)(`n(x) + 2√

dx(`n(y)) =

dx(√x)`n(x) +

√x · d

dx(`n(x))

∴y′

2√x`n(x) +

√x · 1

∴y′

+1√x

∴ y′ =(x√x)(`n(x) + 2√

dx(`n(y)) =

dx(√x)`n(x) +

√x · d

dx(`n(x))

∴y′

2√x`n(x) +

√x · 1

∴y′

+1√x

∴ y′ =(x√x)(`n(x) + 2√

dx(`n(y)) =

dx(√x)`n(x) +

√x · d

dx(`n(x))

∴y′

2√x`n(x) +

√x · 1

∴y′

+1√x

∴ y′ =(x√x)(`n(x) + 2√

Another useful application:

Suppose we have a function like:

y =x3/4√x2 + 1

(3x + 2)5

We could differentiate this thinking of

y = f(x)g(x) , applying the quotient rule, and

then applying the product rule to find f ′(x).

This method will work but it will be very

tough and we are likely to make a mistake

somewhere along the way. So, instead, we

use logarithmic differentiation.

y =x3/4√x2 + 1

(3x + 2)5

y =x3/4√x2 + 1

(3x + 2)5

somewhere along the way.

So, instead, we

y =x3/4√x2 + 1

(3x + 2)5

Logarithmic differentiation

y =x3/4√x2 + 1

(3x + 2)5

1. Take the natural logarithm of both sides

and simplify with Log Laws.

2. Differentiate both sides with respect to x

(implicitly on the LHS).

3. Solve for y′.

It is important to remember that

(ln(g(x))

g′(x)

y =x3/4√x2 + 1

(3x + 2)5

1. Take the natural logarithm of both sides

and simplify with Log Laws.

2. Differentiate both sides with respect to x

(implicitly on the LHS).

3. Solve for y′.

It is important to remember that

(ln(g(x))

g′(x)

y =x3/4√x2 + 1

(3x + 2)5

∴ `n(y) = `n

(x3/4√x2 + 1

(3x + 2)5

(√x2 + 1

)− `n

((3x + 2)5

4`n(x) +

2`n(x2 + 1)− 5`n(3x + 2)

y =x3/4√x2 + 1

(3x + 2)5

∴ `n(y) = `n

(x3/4√x2 + 1

(3x + 2)5

= `n(x3/4

(√x2 + 1

)− `n

((3x + 2)5

4`n(x) +

2`n(x2 + 1)− 5`n(3x + 2)

y =x3/4√x2 + 1

(3x + 2)5

∴ `n(y) = `n

(x3/4√x2 + 1

(3x + 2)5

(√x2 + 1

)− `n

((3x + 2)5

4`n(x) +

2`n(x2 + 1)− 5`n(3x + 2)

y =x3/4√x2 + 1

(3x + 2)5

∴ `n(y) = `n

(x3/4√x2 + 1

(3x + 2)5

(√x2 + 1

)− `n

((3x + 2)5

4`n(x) +

2`n(x2 + 1)− 5`n(3x + 2)

We have

`n(y) =3

4`n(x)+

2`n(x2+1)− 5`n(3x+2)

Now we differentiate both sides w.r.t x

∴y′

x2 + 1

)(2x)−5

3x + 2

∴y′

x2 + 1− 15

3x + 2

∴ y′ =x3/4√x2 + 1

(3x + 2)5

x2 + 1− 15

3x + 2

We have

`n(y) =3

4`n(x)+

2`n(x2+1)− 5`n(3x+2)

∴y′

x2 + 1

)(2x)−5

3x + 2

∴y′

x2 + 1− 15

3x + 2

∴ y′ =x3/4√x2 + 1

(3x + 2)5

x2 + 1− 15

3x + 2

We have

`n(y) =3

4`n(x)+

2`n(x2+1)− 5`n(3x+2)

∴y′

x2 + 1

)(2x)−5

3x + 2

∴y′

x2 + 1− 15

3x + 2

∴ y′ =x3/4√x2 + 1

(3x + 2)5

x2 + 1− 15

3x + 2

We have

`n(y) =3

4`n(x)+

2`n(x2+1)− 5`n(3x+2)

∴y′

x2 + 1

)(2x)−5

3x + 2

∴y′

x2 + 1− 15

3x + 2

∴ y′ =x3/4√x2 + 1

(3x + 2)5

x2 + 1− 15

3x + 2

We have

`n(y) =3

4`n(x)+

2`n(x2+1)− 5`n(3x+2)

∴y′

x2 + 1

)(2x)−5

3x + 2

∴y′

x2 + 1− 15

3x + 2

∴ y′ =x3/4√x2 + 1

(3x + 2)5

x2 + 1− 15

3x + 2

Here are two more examples to which you

can apply logarithmic differentiation.

Attempt them on your own before looking at

the solutions:

1. f (x) =(x + 1)10

(2x− 4)8

2. h(x) =x8 cos3 x√

x− 1

Solution (1.):

`n(f (x)) = `n

((x + 1)10

(2x− 4)8

)∴ `n(f (x)) = `n

((x + 1)10

)− `n

((2x− 4)8

)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)

∴f ′(x)

f (x)=

x + 1− 16

2x− 4

∴ f ′(x) =(x + 1)10

(2x− 4)8

x + 1− 8

x− 2

Solution (1.): `n(f (x)) = `n

((x + 1)10

(2x− 4)8

∴ `n(f (x)) = `n((x + 1)10

)− `n

((2x− 4)8

)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)

∴f ′(x)

f (x)=

x + 1− 16

2x− 4

∴ f ′(x) =(x + 1)10

(2x− 4)8

x + 1− 8

x− 2

((x + 1)10

(2x− 4)8

)∴ `n(f (x)) = `n

((x + 1)10

)− `n

((2x− 4)8

∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)

∴f ′(x)

f (x)=

x + 1− 16

2x− 4

∴ f ′(x) =(x + 1)10

(2x− 4)8

x + 1− 8

x− 2

((x + 1)10

(2x− 4)8

)∴ `n(f (x)) = `n

((x + 1)10

)− `n

((2x− 4)8

)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)

∴f ′(x)

f (x)=

x + 1− 16

2x− 4

∴ f ′(x) =(x + 1)10

(2x− 4)8

x + 1− 8

x− 2

((x + 1)10

(2x− 4)8

)∴ `n(f (x)) = `n

((x + 1)10

)− `n

((2x− 4)8

)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)

∴f ′(x)

f (x)=

x + 1− 16

2x− 4

∴ f ′(x) =(x + 1)10

(2x− 4)8

x + 1− 8

x− 2

((x + 1)10

(2x− 4)8

)∴ `n(f (x)) = `n

((x + 1)10

)− `n

((2x− 4)8

)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)

∴f ′(x)

f (x)=

x + 1− 16

2x− 4

∴ f ′(x) =(x + 1)10

(2x− 4)8

x + 1− 8

x− 2

((x + 1)10

(2x− 4)8

)∴ `n(f (x)) = `n

((x + 1)10

)− `n

((2x− 4)8

)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)

∴f ′(x)

f (x)=

x + 1− 16

2x− 4

∴ f ′(x) =(x + 1)10

(2x− 4)8

x + 1− 8

x− 2

The answer we have below is fine as a final

answer.

f ′(x) =(x + 1)10

(2x− 4)8

x + 1− 8

x− 2

)If you were feeling energetic you could

simplify it further to

f ′(x) =(x + 1)9(x− 14)

128(x− 2)9

Solution (2.):

`n(h(x)) = `n(x8 cos3 x√

)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n

(√x− 1

)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1

2`n(x−1)

∴h′(x)

(− sinx)

cosx− 1

2(x− 1)

∴ h′(x) =

(x8 cos3 x√

x− 1

x− 3 tanx− 1

2(x− 1)

Solution (2.): `n(h(x)) = `n(x8 cos3 x√

∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n(√

x− 1)

∴ `n(h(x)) = 8`n(x)+3`n(cosx)−12`n(x−1)

∴h′(x)

(− sinx)

cosx− 1

2(x− 1)

∴ h′(x) =

(x8 cos3 x√

x− 1

x− 3 tanx− 1

2(x− 1)

)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n

(√x− 1

∴ `n(h(x)) = 8`n(x)+3`n(cosx)−12`n(x−1)

∴h′(x)

(− sinx)

cosx− 1

2(x− 1)

∴ h′(x) =

(x8 cos3 x√

x− 1

x− 3 tanx− 1

2(x− 1)

)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n

(√x− 1

)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1

2`n(x−1)

∴h′(x)

(− sinx)

cosx− 1

2(x− 1)

∴ h′(x) =

(x8 cos3 x√

x− 1

x− 3 tanx− 1

2(x− 1)

)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n

(√x− 1

)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1

2`n(x−1)

∴h′(x)

(− sinx)

cosx− 1

2(x− 1)

∴ h′(x) =

(x8 cos3 x√

x− 1

x− 3 tanx− 1

2(x− 1)

)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n

(√x− 1

)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1

2`n(x−1)

∴h′(x)

(− sinx)

cosx− 1

2(x− 1)

∴ h′(x) =

(x8 cos3 x√

x− 1

x− 3 tanx− 1

2(x− 1)

)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n

(√x− 1

)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1

2`n(x−1)

∴h′(x)

(− sinx)

cosx− 1

2(x− 1)

∴ h′(x) =

(x8 cos3 x√

x− 1

x− 3 tanx− 1

2(x− 1)

These last few slides are here just to give you

a better understanding of the number e.

Previously, we introduced e as follows:

e is the number such that

limh→0

eh − 1

In other words, it is the number such that the

curve y = ex has a gradient of 1 at x = 0.

Now we want to express e as a limit.

limh→0

eh − 1

limh→0

eh − 1

The number e as a limit

Let f (x) = `n(x). We know that f ′(x) =1

xand hence f ′(1) = 1/1 = 1.

Therefore

1 = f ′(1) = limh→0

f (1 + h)− f (1)

= limh→0

`n(1 + h)− `n(1)

= limx→0

`n(1 + x)− `n(1)

= limx→0

x`n(1 + x) = lim

x→0`n(1 + x)1/x

The number e as a limit

Let f (x) = `n(x). We know that f ′(x) =1

xand hence f ′(1) = 1/1 = 1. Therefore

1 = f ′(1) = limh→0

f (1 + h)− f (1)

= limh→0

`n(1 + h)− `n(1)

= limx→0

`n(1 + x)− `n(1)

= limx→0

x`n(1 + x) = lim

x→0`n(1 + x)1/x

e = e1 = elimx→0 `n(1+x)1/x

= limx→0

e`n(1+x)1/x

= limx→0

(1 + x)1/x

That is,

e = limx→0

(1 + x)1/x

As x→ 0, the bracket (1 + x) gets very close

to 1, but the 1/x will become very large.

e = e1 = elimx→0 `n(1+x)1/x

= limx→0

e`n(1+x)1/x

= limx→0

(1 + x)1/x

That is,

e = limx→0

(1 + x)1/x

e = e1 = elimx→0 `n(1+x)1/x

= limx→0

e`n(1+x)1/x

= limx→0

(1 + x)1/x

That is,

e = limx→0

(1 + x)1/x

We have that e = limx→0

(1 + x)1/x.

Here are some calculations for x→ 0+.

Prescribed tut problems:

Complete the following exercises from the

8th edition:

Ch 3.6:

2, 5, 8, 9, 13, 20, 29, 34, 39, 41, 44, 49, 52

If you are using the 7th edition:

Ch 3.6: 13 → 11

mat01a1: derivatives of log functions and logarithmic … · mat01a1: derivatives of log functions...

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