3.1 Derivative of a Function

0

limh

f a h f ah

is called the derivative of at .f a

We write: 0

limh

f a h f af x

h

“The derivative of f with respect to x is …”

There are many ways to write the derivative of y f x

3.1 Derivative of a Function

f x “f prime x” or “the derivative of f with respect to x”

y “y prime”

dydx “dee why dee ecks” or “the derivative of y with

respect to x”dfdx “dee eff dee ecks” or “the derivative of f with

respect to x”

d f xdx “dee dee ecks uv eff uv ecks” or “the derivative

of f of x”( of of )d dx f x


dx does not mean d times x !

dy does not mean d times y !


dydx does not mean !dy dx

(except when it is convenient to think of it as division.)

dfdx

does not mean !df dx

(except when it is convenient to think of it as division.)


(except when it is convenient to treat it that way.)

d f xdx

does not mean times !ddx

f x


The derivative is the slope of the original function.

The derivative is defined at the end points of a function on a closed interval.

y f x

y f x


2 3y x

2 2

0

3 3limh

x h xy

h

2y x

0lim 2h

y x h


A function is differentiable if it has a derivative everywhere in its domain. It must be continuous and smooth. Functions on closed intervals must have one-sided derivatives defined at the end points.


To be differentiable, a function must be continuous and smooth.Derivatives will fail to exist at:

corner

f x x

cusp

23f x x

vertical tangent 3f x x

discontinuity

1, 0 1, 0

xf x

x

3.2 Differentiability

Most of the functions we study in calculus will be differentiable.


There are two theorems on page 110:

If f has a derivative at x = a, then f is continuous at x = a.

Since a function must be continuous to have a derivative, if it has a derivative then it is continuous.


12

f a

3f b

Intermediate Value Theorem for Derivatives

Between a and b, must take

on every value between and .

f 12 3

If a and b are any two points in an interval on which f is

differentiable, then takes on every value between

and .

f f a

f b


If the derivative of a function is its slope, then for a constant function, the derivative must be zero.

0d cdx

example: 3y

0y

The derivative of a constant is zero.

3.3 Rules for Differentiation

We saw that if , .2y x 2y x

This is part of a pattern.

1n nd x nxdx

examples:

4f x x

34f x x

8y x

78y x

power rule


1n nd x nxdx


Proof:

hxhxx

dxd nn

h

n

)(lim0

hxhhnxxx

dxd nnnn

h

n

...lim1

0

hhhnxx

dxd nn

h

n

...lim1

0

1

0lim

n

h

n nxxdxd

d ducu cdx dx

examples:

1n nd cx cnxdx

constant multiple rule:

5 4 47 7 5 35d x x xdx


(Each term is treated separately)

d ducu cdx dx

constant multiple rule:

sum and difference rules:

d du dvu vdx dx dx

d du dvu vdx dx dx

4 12y x x 34 12y x

4 22 2y x x

34 4dy x xdx


Find the horizontal tangents of: 4 22 2y x x 34 4dy x x

dx

Horizontal tangents occur when slope = zero.34 4 0x x

3 0x x

2 1 0x x

1 1 0x x x

0, 1, 1x

Substituting the x values into the original equation, we get:

2, 1, 1y y y (The function is even, so we only get two horizontal tangents.)


4 22 2y x x

2y

1y


4 22 2y x x

First derivative (slope) is zero at:

0, 1, 1x

34 4dy x xdx


product rule: d dv duuv u v

dx dx dx Notice that this is not just the

product of two derivatives.

This is sometimes memorized as: d uv u dv v du

2 33 2 5d x x xdx

5 3 32 5 6 15d x x x xdx

5 32 11 15d x x xdx

4 210 33 15x x

2 3x 26 5x 32 5x x 2x

4 2 2 4 26 5 18 15 4 10x x x x x

4 210 33 15x x


product rule: d dv duuv u v

dx dx dx


Proof

hxvxuhxvhxuuv

dxd

h

)()()()(lim)(0

add and subtract u(x+h)v(x)in the denominator

hxvhxuxvhxuxvxuhxvhxuuv

dxd

h

)()()()()()()()(lim)(0

h

xuhxuxvxvhxvhxuuvdxd

h

)()()()()()(lim)(0

dxduv

dxdvuuv

dxd

)(

quotient rule:

2

du dvv ud u dx dxdx v v

or 2

u v du u dvdv v

3

2

2 53

d x xdx x

2 2 3

22

3 6 5 2 5 2

3

x x x x x

x


Higher Order Derivatives:dyydx

is the first derivative of y with respect to x.

2

2

dy d dy d yydx dx dx dx

is the second derivative.(y double prime)

dyydx

is the third derivative.

4 dy ydx

is the fourth derivative.

We will learn later what these higher order derivatives are used for.



Suppose u and v are functions that are differentiable atx = 3, and that u(3) = 5, u’(3) = -7, v(3) = 1, and v’(3)= 4.Find the following at x = 3 :

)(.1 uvdxd

'')( vuuvuvdxd

8)7)(1()3(5

vu

dxd.2

2

''v

uvvuvu

dxd

21)4)(5()7)(1( 27

uv

dxd.3

2

''u

vuuvuv

dxd

25)7)(1()4)(5(

2527


hiho

dxd

))(()()()()(

hohohidhohodhi


Consider a graph of displacement (distance traveled) vs. time.

time (hours)

distance(miles)

Average velocity can be found by taking:change in position

change in timest

t

sA

B

ave

f t t f tsVt t

The speedometer in your car does not measure average velocity, but instantaneous velocity.

0

limt

f t t f tdsV tdt t

(The velocity at one moment in time.)

3.4 Velocity and other Rates of Change


Velocity is the first derivative of position.

Acceleration is the second derivative of position.

Example: Free Fall Equation

21 2

s g t

GravitationalConstants:

2

ft32 sec

g

2

m9.8 sec

g

2

cm980 sec

g

21 32 2

s t

216 s t 32 dsV tdt

Speed is the absolute value of velocity.


Acceleration is the derivative of velocity.

dvadt

2

2

d sdt

example: 32v t

32a If distance is in: feet

Velocity would be in:feetsec

Acceleration would be in:

ftsec sec

2

ftsec


time

distance

acc posvel pos &increasing

acc zerovel pos &constant

acc negvel pos &decreasing

velocityzero

acc negvel neg &decreasing acc zero

vel neg &constant

acc posvel neg &increasing

acc zero,velocity zero


Rates of Change:

Average rate of change = f x h f x

h

Instantaneous rate of change = 0

limh

f x h f xf x

h

These definitions are true for any function.

( x does not have to represent time. )


For a circle: 2A r2dA d r

dr dr

2dA rdr

Instantaneous rate of change of the area withrespect to the radius.

For tree ring growth, if the change in area is constant then dr must get smaller as r gets larger.

2 dA r dr


from Economics:

Marginal cost is the first derivative of the cost function, and represents an approximation of the cost of producing one more unit.


Example 13: Suppose it costs: 3 26 15c x x x x

to produce x stoves. 23 12 15c x x x

If you are currently producing 10 stoves, the 11th stove will cost approximately:

210 3 10 12 10 15c 300 120 15

$195marginal cost

The actual cost is: 11 10C C

3 2 3 211 6 11 15 11 10 6 10 15 10

770 550 $220 actual cost


Note that this is not a great approximation – Don’t let that bother you.

Marginal cost is a linear approximation of a curved function. For large values it gives a good approximation of the cost of producing the next item.



20

2

Consider the function siny

We could make a graph of the slope: slope

1

0

1

0

1Now we connect the dots!The resulting curve is a cosine curve.

sin cosd x xdx

3.5 Derivatives of Trigonometric Functions


hxhxx

dxd

h

sin)sin(limsin0

hxxhhxx

dxd

h

sincossincossinlimsin0

hxh

hhxx

dxd

hh

cossinlim)1(cossinlimsin00

hxhhxx

dxd

h

cossin)1(cossinlimsin0

Proof


hxh

hhxx

dxd

hh

cossinlim)1(cossinlimsin00

= 0 = 1

sin cosd x xdx


hxhxx

dxd

h

cos)cos(limcos0

hxxhhxx

dxd

h

cossinsincoscoslimcos0

hxh

hhxx

dxd

hh

sinsinlim)1(coscoslimcos00

hxhhxx

dxd

h

sinsin)1(coscoslimcos0

Find the derivative of cos x


= 0 = 1

hxh

hhxx

dxd

hh

sinsinlim)1(coscoslimcos00

cos sind x xdx

We can find the derivative of tangent x by using the quotient rule.

tand xdx

sincos

d xdx x

2

cos cos sin sincos

x x x xx

2 2

2

cos sincosx x

x

2

1cos x

2sec x

2tan secd x xdx


Derivatives of the remaining trig functions can be determined the same way.

sin cosd x xdx

cos sind x xdx

2tan secd x xdx

2cot cscd x xdx

sec sec tand x x xdx

csc csc cotd x x xdx



Jerk A sudden change in acceleration

Definition JerkJerk is the derivative of acceleration. If a body’s positionat time t is s(t), the body’s jerk at time t is

3

3

2

2

)(dt

sddt

vddtdatj


Consider a simple composite function:6 10y x

2 3 5y x

If 3 5u x

then 2y u

6 10y x 2y u 3 5u x

6dydx 2dy

du 3du

dx

dy dy dudx du dx

6 2 3

3.6 Chain Rule

dy dy dudx du dx Chain Rule:

example: sinf x x 2 4g x x Find: at 2f g x

cosf x x 2g x x 2 4 4 0g

0 2f g cos 0 2 2 1 4 4

3.6 Chain Rule

If is the composite of and , then:

f g y f u u g x

at at xu g xf g f g )('))((' xgxgf

2sin 4f g x x

2sin 4y x

siny u 2 4u x

cosdy udu 2du x

dx

dy dy dudx du dx

cos 2dy u xdx

2cos 4 2dy x xdx

2cos 2 4 2 2dydx

cos 0 4dydx

4dydx

3.6 Chain Rule

Here is a faster way to find the derivative:

2sin 4y x

2 2cos 4 4dy x xdx

2cos 4 2y x x

Differentiate the outside function...

…then the inside function

At 2, 4x y

3.6 Chain Rule

2cos 3d xdx

2cos 3d x

dx

2 cos 3 cos 3dx xdx

2cos 3 sin 3 3dx x xdx

2cos 3 sin 3 3x x

6cos 3 sin 3x x

The chain rule can be used more than once.

(That’s what makes the “chain” in the “chain rule”!)

3.6 Chain Rule

Derivative formulas include the chain rule!

1n nd duu nudx dx

sin cosd duu udx dx

cos sind duu udx dx

2tan secd duu udx dx

etcetera…

3.6 Chain Rule

3.6 Chain RuleFind

)3cos( 2 xxy )16)(3sin( 2 xxxdxdy

))sin(cos(xy

)24(cos 33 xxy

)sin)(cos(cos xxdxdy

)212))(24sin()(24(cos3 2332 xxxxxdxdy

))24sin()(24(cos)636( 3322 xxxxxdxdy

dxdy

The chain rule enables us to find the slope of parametrically defined curves:

dy dy dxdt dx dt

dydydt

dx dxdt

The slope of a parametrized curve is given by:

dydy dt

dxdxdt

3.6 Chain Rule

These are the equations for an ellipse.

Example:

3cosx t 2siny t

3sindx tdt 2cosdy t

dt

2cos3sin

dy tdx t

2 cot3

t

3.6 Chain Rule

2 2 1x y This is not a function, but it would still be nice to be able to find the slope.

2 2 1d d dx ydx dx dx

Do the same thing to both sides.

2 2 0dyx ydx

Note use of chain rule.

2 2dyy xdx

22

dy xdx y

dy x

dx y

3.7 Implicit Differentiation

22 siny x y

22 sind d dy x ydx dx dx

This can’t be solved for y.

2 2 cosdy dyx ydx dx

2 cos 2dy dyy xdx dx

22 cosdy xydx

22 cos

dy xdx y

This technique is called implicit differentiation.

1 Differentiate both sides w.r.t. x.

2 Solve for .dydx


3.7 Implicit DifferentiationImplicit Differentiation Process

1. Differentiate both sides of the equation with respect to x.2. Collect the terms with dy/dx on one side of the equation.3. Factor out dy/dx .4. Solve for dy/dx .

Find the equations of the lines tangent and normal to the

curve at .2 2 7x xy y ( 1, 2)2 2 7x xy y

2 2 0dydyx yx ydxdx

Note product rule.

2 2 0dy dyx x y ydx dx

22dy y xy xdx

22

dy y xdx y x

2 2 12 2 1

m

2 24 1

45


Find the equations of the lines tangent and normal to the

curve at .2 2 7x xy y ( 1, 2)

45

mtangent:

42 15

y x

4 425 5

y x

4 145 5

y x

normal:

52 14

y x

5 524 4

y x

5 34 4

y x



Find if .2

2

d ydx

3 22 3 7x y

3 22 3 7x y

26 6 0x y y

26 6y y x 26

6xyy

2xyy

2

2

2y x x yyy

2

2

2x xy yy y

2 2

2

2x xyy

xyy

4

3

2x xyy y

Substitute back into the equation.

y


3.7 Implicit DifferentiationRational Powers of Differentiable Functions

Power Rule for Rational Powers of x

If n is any rational number, then

1 nn nxx

dxd

3.7 Implicit DifferentiationProof: Let p and q be integers with q > 0.

qp

xy

pq xy

Raise both sides to the q power

Differentiate with respect to x

11 pq pxdxdyqy Solve for dy/dx


1

1

q

p

qypx

dxdy Substitute for y

1/

1

)(

qqp

p

xqpx

dxdy Remove parenthesis

qpp

p

qxpx

dxdy

/

1

Subtract exponents

qpx

dxdy qppp )/(1

1)/( qpx

qp

dxdy

Because x and y are reversed to find the reciprocal function, the following pattern always holds:

2y x

y x

4m 2,4

4,2 14

m

Slopes are reciprocals.

Derivative Formula for Inverses:

dfdx df

dxx f a

x a

1 1

( )

evaluated at ( )f a

is equal to the reciprocal ofthe derivative of ( )f xevaluated at .a

The derivative of 1( )f x

3.8 Derivatives of Inverse Trigonometric Functions

siny x

1siny xWe can use implicit differentiation to find:

1sind xdx

1siny x

sin y x sind dy xdx dx

cos 1dyydx

1cos

dydx y


We can use implicit differentiation to find:

1sind xdx

1siny x

sin y x sind dy xdx dx

cos 1dyydx

1cos

dydx y

2 2sin cos 1y y

2 2cos 1 siny y 2cos 1 siny y

But 2 2y

so is positive.cos y

2cos 1 siny y

2

1

1 sin

dydx y

2

1

1

dydx x


1siny x

1cos

dydx y


)cos(sin1

1 xdxdy

x1sin x

1

21 x211

xdxdy

xy sin

1cos dxdyy


)(tansec1

12 xdxdy

x1tan x

1

21 x

211xdx

dy

xy tan

1sec2 dxdyy

Find xdxd 1tan

xy 1tan

ydxdy

2sec1


)tan(sec)sec(sec1

11 xxdxdy

x1sec

x

1

12 x

1||

12

xxdx

dyxy sec

1tansec dxdyyy

Find xdxd 1sec

xy 1sec

yydxdy

tansec1

1

2

1sin1

d duudx dxu

12

1tan1

d duudx u dx

1

2

1sec1

d duudx dxu u

1

2

1cos1

d duudx dxu

12

1cot1

d duudx u dx

1

2

1csc1

d duudx dxu u

1 1cos sin2

x x 1 1cot tan2

x x 1 1csc sec2

x x


Your calculator contains all six inverse trig functions.However it is occasionallystill useful to know the following:

1 1 1sec cosxx

1 1cot tan2

x x

1 1 1csc sinxx



Find

)3(cos 21 xy 422 916)6(

)3(1(1

xxx

xdxdy

xy 1cot 1

xxy 1sec

111

11

122

2

xxx

dxdy

)1)((sec1||

1 1

2x

xxx

dxdy

dxdy

Look at the graph of xy e

The slope at x = 0 appears to be 1.

If we assume this to be true, then:

0 0

0lim 1

h

h

e eh

definition of derivative

3.9 Derivatives of Exponential and Logarithmic Functions

Now we attempt to find a general formula for the derivative of using the definition.xy e

0

limx h x

x

h

d e eedx h

0lim

x h x

h

e e eh

0

1limh

x

h

eeh

0

1limh

x

h

eeh

1xe xe

This is the slope at x = 0, which we have assumed to be 1.


x xd e edx

xe is its own derivative!

If we incorporate the chain rule: u ud due edx dx

We can now use this formula to find the derivative of xa


xd adx

ln xad edx

lnx ad edx

ln lnx a de x adx

Incorporating the chain rule:

lnu ud dua a adx dx


aaadxd xx ln

So far today we have:

u ud due edx dx

lnu ud dua a adx dx

Now it is relatively easy to find the derivative of .ln x


lny xye x

yd de xdx dx

1y dyedx

1y

dydx e

1lnd xdx x

1lnd duudx u dx


To find the derivative of a common log function, you could just use the change of base rule for logs:

logd xdx

lnln10

d xdx

1 ln

ln10d xdx

1 1

ln10 x

The formula for the derivative of a log of any base other than e is:

1loglna

d duudx u a dx


u ud due edx dx

lnu ud dua a adx dx

1loglna

d duudx u a dx

1lnd duu

dx u dx



xey 22

3xy

3ln xy

)(sin 41 xey

Find y’

xey 22'

)2)(3ln(3'2

xy x

xx

xy 3)3(1' 2

3

)4)(()(1

1' 4

24

x

xe

ey


Logarithmic differentiation

Used when the variable is in the base and the exponent

y = xx

ln y = ln xx

ln y = x ln x

xx

xdxdy

yln11

xydxdy ln1

xxdxdy x ln1

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3.1 Derivative of a Function