Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)

..

Sec on 3.3Deriva ves of Logarithmic and

Exponen al Func ons

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

March 21, 2011

Announcements

I Quiz 3 next week on 2.6,2.8, 3.1, 3.2

ObjectivesI Know the deriva ves of theexponen al func ons (with anybase)

I Know the deriva ves of thelogarithmic func ons (with anybase)

I Use the technique of logarithmicdifferen a on to find deriva vesof func ons involving roducts,quo ents, and/or exponen als.

OutlineRecall Sec on 3.1–3.2

Deriva ve of the natural exponen al func onExponen al Growth

Deriva ve of the natural logarithm func on

Deriva ves of other exponen als and logarithmsOther exponen alsOther logarithms

Logarithmic Differen a onThe power rule for irra onal powers

Conventions on power expressionsLet a be a posi ve real number.

I If n is a posi ve whole number, then an = a · a · · · · · a︸︷︷︸n factors

I a0 = 1.I For any real number r, a−r =

1ar .

I For any posi ve whole number n, a1/n = n√a.

There is only one con nuous func on which sa sfies all of theabove. We call it the exponen al func on with base a.

Properties of exponentialsTheoremIf a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on withdomain (−∞,∞) and range (0,∞). In par cular, ax > 0 for all x.For any real numbers x and y, and posi ve numbers a and b we have

I ax+y = axay

I ax−y =ax

ay

(nega ve exponents mean reciprocals)

I (ax)y = axy

(frac onal exponents mean roots)

I (ab)x = axbx


I ax+y = axay

I ax−y =ax

ay (nega ve exponents mean reciprocals)

I (ax)y = axy

(frac onal exponents mean roots)

I (ab)x = axbx


I ax+y = axay

I ax−y =ax

ay (nega ve exponents mean reciprocals)

I (ax)y = axy (frac onal exponents mean roots)I (ab)x = axbx

Graphs of exponential functions

.. x.

y

.y = 1x

.

y = 2x

.

y = 3x

.

y = 10x

.

y = 1.5x

.

y = (1/2)x

.

y = (1/3)x

.

y = (1/10)x

.

y = (2/3)x

The magic number

Defini on

e = limn→∞

(1+

1n

)n

= limh→0+

(1+ h)1/h

Existence of eSee Appendix B

I We can experimentallyverify that this numberexists and is

e ≈ 2.718281828459045 . . .

I e is irra onalI e is transcendental

n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828

Logarithms

Defini on

I The base a logarithm loga x is the inverse of the func on ax

y = loga x ⇐⇒ x = ay

I The natural logarithm ln x is the inverse of ex. Soy = ln x ⇐⇒ x = ey.

Facts about Logarithms

Facts

(i) loga(x1 · x2) = loga x1 + loga x2

(ii) loga

(x1x2

)= loga x1 − loga x2

(iii) loga(xr) = r loga x

Graphs of logarithmic functions

.. x.

y

.

y = 2x

.

y = log2 x

..

(0, 1)

..(1, 0).

y = 3x

.

y = log3 x

.

y = 10x

.y = log10 x.

y = ex

.

y = ln x

Change of base formula

FactIf a > 0 and a ̸= 1, and the same for b, then

loga x =logb xlogb a

Upshot of changing baseThe point of the change of base formula

loga x =logb xlogb a

=1

logb a· logb x = (constant) · logb x

is that all the logarithmic func ons are mul ples of each other. Sojust pick one and call it your favorite.

I Engineers like the common logarithm log = log10I Computer scien sts like the binary logarithm lg = log2I Mathema cians like natural logarithm ln = loge

Naturally, we will follow the mathema cians. Just don’t pronounceit “lawn.”






Derivatives of ExponentialsFactIf f(x) = ax, then f′(x) = f′(0)ax.

Proof.Follow your nose:

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

ax+h − ax

h

= limh→0

axah − ax

h= ax · lim

h→0

ah − 1h

= ax · f′(0).

Derivatives of ExponentialsFactIf f(x) = ax, then f′(x) = f′(0)ax.

Proof.Follow your nose:

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

ax+h − ax

h

= limh→0

axah − ax

h= ax · lim

h→0

ah − 1h

= ax · f′(0).

The funny limit in the case of eQues on

What is limh→0

eh − 1h

?

Solu on


What is limh→0

eh − 1h

?

Solu on

Recall e = limn→∞

(1+

1n

)n

= limh→0

(1+ h)1/h. If h is small enough,

e ≈ (1+ h)1/h. So

eh − 1h

≈[(1+ h)1/h

]h − 1h

=(1+ h)− 1

h=

hh= 1


What is limh→0

eh − 1h

?

Solu onSo in the limit we get equality:

limh→0

eh − 1h

= 1

Derivative of the naturalexponential function

From

ddx

ax =

(limh→0

ah − 1h

)ax and lim

h→0

eh − 1h

= 1

we get:

Theorem

ddx

ex = ex

Exponential GrowthI Commonly misused term to say something grows exponen allyI It means the rate of change (deriva ve) is propor onal to thecurrent value

I Examples: Natural popula on growth, compounded interest,social networks

Examples

Example

Findddx

e3x.

Solu on

ddx

e3x = e3xddx

(3x) = 3e3x

Examples

Example

Findddx

e3x.

Solu on

ddx

e3x = e3xddx

(3x) = 3e3x

Examples

Example

Findddx

ex2.

Solu on

ddx

ex2= ex

2 ddx

(x2) = 2xex2

Examples

Example

Findddx

ex2.

Solu on

ddx

ex2= ex

2 ddx

(x2) = 2xex2

Examples

Example

Findddx

x2ex.

Solu on

ddx

x2ex = 2xex + x2ex

Examples

Example

Findddx

x2ex.

Solu on

ddx

x2ex = 2xex + x2ex






Derivative of the natural logarithmLet y = ln x. Then x = ey so

eydydx

= 1

=⇒ dydx

=1ey

=1x

We have discovered:Fact

ddx

ln x =1x

.. x.

y

.

ln x

.

1x


eydydx

= 1

=⇒ dydx

=1ey

=1x


ddx

ln x =1x

.. x.

y

.

ln x

.

1x


eydydx

= 1

=⇒ dydx

=1ey

=1x


ddx

ln x =1x

.. x.

y

.

ln x

.

1x


eydydx

= 1

=⇒ dydx

=1ey

=1x


ddx

ln x =1x

.. x.

y

.

ln x

.

1x


eydydx

= 1

=⇒ dydx

=1ey

=1x


ddx

ln x =1x

.. x.

y

.

ln x

.

1x


eydydx

= 1

=⇒ dydx

=1ey

=1x


ddx

ln x =1x

.. x.

y

.

ln x

.

1x

The Tower of Powersy y′

x3 3x2

x2 2x1

x1 1x0

x0 0

? ?

x−1 −1x−2

x−2 −2x−3

I The deriva ve of a power func on is apower func on of one lower power

I Each power func on is the deriva ve ofanother power func on, except x−1

I ln x fills in this gap precisely.


x3 3x2

x2 2x1

x1 1x0

x0 0

? x−1

x−1 −1x−2

x−2 −2x−3





x3 3x2

x2 2x1

x1 1x0

x0 0

ln x x−1

x−1 −1x−2

x−2 −2x−3




Examples

Examples

Find deriva ves of these func ons:I ln(3x)I x ln xI ln

√x

ExamplesExample

Findddx

ln(3x).

ExamplesExample

Findddx

ln(3x).

Solu on (chain rule way)

ddx

ln(3x) =13x

· 3 =1x

ExamplesExample

Findddx

ln(3x).

Solu on (proper es of logarithms way)

ddx

ln(3x) =ddx

(ln(3) + ln(x)) = 0+1x=

1x

The first answermight be surprising un l you see the second solu on.

ExamplesExample

Findddx

x ln x

Solu onThe product rule is in play here:

ddx

x ln x =(

ddx

x)ln x+ x

(ddx

ln x)

= 1 · ln x+ x · 1x= ln x+ 1

ExamplesExample

Findddx

x ln x

Solu onThe product rule is in play here:

ddx

x ln x =(

ddx

x)ln x+ x

(ddx

ln x)

= 1 · ln x+ x · 1x= ln x+ 1

ExamplesExample

Findddx

ln√x.

ExamplesExample

Findddx

ln√x.

Solu on (chain rule way)

ddx

ln√x =

1√xddx

√x =

1√x

12√x=

12x

ExamplesExample

Findddx

ln√x.

Solu on (proper es of logarithms way)

ddx

ln√x =

ddx

(12ln x

)=

12ddx

ln x =12· 1x

The first answermight be surprising un l you see the second solu on.






Other logarithmsExample

Use implicit differen a on to findddx

ax.

Solu onLet y = ax, so

ln y = ln ax = x ln a

Differen ate implicitly:

1ydydx

= ln a =⇒ dydx

= (ln a)y = (ln a)ax



ax.




1ydydx

= ln a =⇒ dydx




ax.




1ydydx

= ln a =⇒ dydx


The funny limit in the case of aLet y = ex. Before we showed y′ = y′(0)y, and now we knowy′ = (ln a)y. So

Corollary

limh→0

ah − 1h

= ln a

In par cular

ln 2 = limh→0

2h − 1h

≈ 0.693 ln 3 = limh→0

3h − 1h

≈ 1.10


Findddx

loga x.

Solu on


Findddx

loga x.

Solu onLet y = loga x, so ay = x.

Now differen ate implicitly:

(ln a)aydydx

= 1 =⇒ dydx

=1

ay ln a=

1x ln a


Findddx

loga x.

Solu onLet y = loga x, so ay = x. Now differen ate implicitly:

(ln a)aydydx

= 1 =⇒ dydx

=1

ay ln a=

1x ln a


Findddx

loga x.

Solu onOr we can use the change of base formula:

y =ln xln a

=⇒ dydx

=1ln a

1x

More examplesExample

Findddx

log2(x2 + 1)

Answer

dydx

=1ln 2

1x2 + 1

(2x) =2x

(ln 2)(x2 + 1)

More examplesExample

Findddx

log2(x2 + 1)

Answer

dydx

=1ln 2

1x2 + 1

(2x) =2x

(ln 2)(x2 + 1)






A nasty derivativeExample

Let y =(x2 + 1)

√x+ 3

x− 1. Find y′.

Solu onWe use the quo ent rule, and the product rule in the numerator:

y′ =(x− 1)

[2x√x+ 3+ (x2 + 1)12(x+ 3)−1/2

]− (x2 + 1)

√x+ 3(1)

(x− 1)2

=2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2

A nasty derivativeExample

Let y =(x2 + 1)

√x+ 3

x− 1. Find y′.

Solu onWe use the quo ent rule, and the product rule in the numerator:

y′ =(x− 1)

[2x√x+ 3+ (x2 + 1)12(x+ 3)−1/2

]− (x2 + 1)

√x+ 3(1)

(x− 1)2

=2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2

Another way

y =(x2 + 1)

√x+ 3

x− 1

ln y = ln(x2 + 1) +12ln(x+ 3)− ln(x− 1)

1ydydx

=2x

x2 + 1+

12(x+ 3)

− 1x− 1

So

dydx

=

(2x

x2 + 1+

12(x+ 3)

− 1x− 1

)y

Another way

y =(x2 + 1)

√x+ 3

x− 1

ln y = ln(x2 + 1) +12ln(x+ 3)− ln(x− 1)

1ydydx

=2x

x2 + 1+

12(x+ 3)

− 1x− 1

So

dydx

=

(2x

x2 + 1+

12(x+ 3)

− 1x− 1

)y

Another way

y =(x2 + 1)

√x+ 3

x− 1

ln y = ln(x2 + 1) +12ln(x+ 3)− ln(x− 1)

1ydydx

=2x

x2 + 1+

12(x+ 3)

− 1x− 1

So

dydx

=

(2x

x2 + 1+

12(x+ 3)

− 1x− 1

)y

Another way

y =(x2 + 1)

√x+ 3

x− 1

ln y = ln(x2 + 1) +12ln(x+ 3)− ln(x− 1)

1ydydx

=2x

x2 + 1+

12(x+ 3)

− 1x− 1

So

dydx

=

(2x

x2 + 1+

12(x+ 3)

− 1x− 1

)y

Another way

y =(x2 + 1)

√x+ 3

x− 1

ln y = ln(x2 + 1) +12ln(x+ 3)− ln(x− 1)

1ydydx

=2x

x2 + 1+

12(x+ 3)

− 1x− 1

So

dydx

=

(2x

x2 + 1+

12(x+ 3)

− 1x− 1

)(x2 + 1)

√x+ 3

x− 1

Compare and contrastI Using the product, quo ent, and power rules:

y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2

I Using logarithmic differen a on:

y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)

I Are these the same?

Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)


Yes.


y′ =2x√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

(x− 1)

I Are these the same? Yes.

Derivatives of powers

Ques on

Let y = xx. Which of these is true?(A) Since y is a power func on,

y′ = x · xx−1 = xx.(B) Since y is an exponen al

func on, y′ = (ln x) · xx

(C) Neither ..x

.

y

..1

..

1

Derivatives of powers

Ques on

Let y = xx. Which of these is true?(A) Since y is a power func on,

y′ = x · xx−1 = xx.(B) Since y is an exponen al

func on, y′ = (ln x) · xx

(C) Neither ..x

.

y

..1

..

1

Why not?Answer

(A) y′ ̸= xx because xx > 0 for allx > 0, and this func ondecreases at some places

(B) y′ ̸= (ln x)xx because (ln x)xx = 0when x = 1, and this func ondoes not have a horizontaltangent at x = 1.

..x

.

y

..1

..

1

Why not?Answer

(A) y′ ̸= xx because xx > 0 for allx > 0, and this func ondecreases at some places

(B) y′ ̸= (ln x)xx because (ln x)xx = 0when x = 1, and this func ondoes not have a horizontaltangent at x = 1.

..x

.

y

..1

..

1

It’s neither!Solu onIf y = xx, then

ln y = x ln x1ydydx

= x · 1x+ ln x = 1+ ln x

dydx

= (1+ ln x)xx = xx + (ln x)xx


ln y = x ln x

1ydydx

= x · 1x+ ln x = 1+ ln x

dydx



ln y = x ln x1ydydx

= x · 1x+ ln x = 1+ ln x

dydx



ln y = x ln x1ydydx

= x · 1x+ ln x = 1+ ln x

dydx


Or both?Solu on

ddx

xx = xx + (ln x)xx = (1+ ln x)xx

Remarks

I Each of these terms is one of thewrong answers!

I y′ < 0 on the interval (0, e−1)

I y′ = 0 when x = e−1

..x

.

y

..1

..

1

Or both?Solu on

ddx


Remarks



I y′ = 0 when x = e−1

..x

.

y

..1

..

1

Or both?Solu on

ddx


Remarks



I y′ = 0 when x = e−1

..x

.

y

..1

..

1

Or both?Solu on

ddx


Remarks



I y′ = 0 when x = e−1

..x

.

y

..1

..

1

Or both?Solu on

ddx


Remarks



I y′ = 0 when x = e−1

..x

.

y

..1

..

1

Or both?Solu on

ddx


Remarks



I y′ = 0 when x = e−1

..x

.

y

..1

..

1

Or both?Solu on

ddx


Remarks



I y′ = 0 when x = e−1

..x

.

y

..1

..

1

Derivatives of power functionswith any exponent

Fact (The power rule)

Let y = xr. Then y′ = rxr−1.

Proof.

y = xr =⇒ ln y = r ln x

Now differen ate:

1ydydx

=rx

=⇒ dydx

= ryx= rxr−1

Derivatives of power functionswith any exponent

Fact (The power rule)

Let y = xr. Then y′ = rxr−1.

Proof.

y = xr =⇒ ln y = r ln x

Now differen ate:

1ydydx

=rx

=⇒ dydx

= ryx= rxr−1

SummaryI Deriva ves ofLogarithmic andExponen al Func ons

I LogarithmicDifferen a on can allowus to avoid the productand quo ent rules.

I We are finally done withthe Power Rule!

y y′

ex ex

ax (ln a) · ax

ln x1x

loga x1ln a

· 1x

Technology

Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)