Index FAQ The derivative as the slope of the tangent line (at a point)

Index FAQ

The derivative as the slope of the tangent line

(at a point)

Index FAQ

Video help: MIT!!! http://ocw.mit.edu/courses/

mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/

Index FAQ

What is a derivative?

A function, which gives the:the rate of change of a

function in generalthe slope of the line tangent

to the curve in general

Index FAQ

What is a differential quotient?

Just a number!the rate of change of a function at a

given pointthe slope of the line tangent to the

curve at a certain pointThe substitutional value of the

derivative

Index FAQ

The tangent line

single pointof intersection

Index FAQ

slope of a secant line

ax

f(x)

f(a)

f(a) - f(x)

a - x

Index FAQ

slope of a (closer) secant line

ax

f(x)

f(a)

f(a) - f(x)

a - x

x

Index FAQ

closer and closer…

a

Index FAQ

watch the slope...

Index FAQ

watch what x does...

ax

Index FAQ

The slope of the secant line gets closer and closer to the slope of the tangent line...

Index FAQ

As the values of x get closer and closer to a!

ax

Index FAQ

The slope of the secant lines gets closer

to the slope of the tangent line...

...as the values of x get closer to a

Translates to….

Index FAQ

limax

f(x) - f(a)x - a

Equation for the slope

Which gives us the the exact slope of the line tangent to the curve at a!

as x goes to a

Differential quotient

Index FAQ

Differential quotient: other form

aa+h

f(a+h)

f(a)

f(x+h) - f(x)

(x+h) - x= f(x+h) - f(x)

h

(For this particular curve, h is a negative value)

h

limh0

Index FAQ

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.

Velocity and other Rates of Change

Index FAQ

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 time

s

t

t

sA

B

ave

f t t f tsV

t t

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

0

limt

f t t f tdsV t

dt t

(The velocity at one moment in time.)

Velocity and other Rates of Change- physical menaing of the differential quotient

Index FAQ

Velocity and other rates of change

Velocity is the first derivative of position.

Acceleration is the second derivative of position.

Index FAQ

Example:Free Fall Equation

21

2s g t

GravitationalConstants:

2

ft32

secg

2

m9.8

secg

2

cm980

secg

2132

2s t

216 s t 32 ds

V tdt

Speed is the absolute value of velocity.

Velocity

Index FAQ

Acceleration is the derivative of velocity.

dva

dt

2

2

d s

dt example:

32v t

32a If distance is in: feet

Velocity would be in:feet

sec

Acceleration would be in:

ftsec sec

2

ft

sec

3.4 Velocity and other Rates of Change

Index FAQtime

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


Index FAQ

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

corner

f x x

cusp

2

3f x x

vertical tangent

3f x x

discontinuity

1, 0

1, 0

xf x

x

Differentiability

Index FAQ

Theorem : f is differentiable on the interval (a,b). f is continuous on the interval (a,b).

Proof: Assume that f ’(c) exists for any c in (a,b).

Then lim [ f(c+h)- f(c)] . h0

= f ’(c) • 0 = 0

f ’(c) = lim f(c+h) -f(c) . h 0 h

= f ’(c) • lim h . h 0

So lim [ f(c+h) - f(c)] = 0 . h0

, and from here we get lim f(c+h) = f(c) . . h0

So f is continuous at c for every c in (a,b).

/ • h

Index FAQ

Example: Since the derivative of f(x)= 5x2+x+1 is f ’(x) = 10x+1, which exists for every real number x. So f(x)= 5x2+x+1 is continuous everywhere.

RemarkThe reverse of this theorem is not

true.

Counter example: We know that f(x) = |x| is continuous on R , but at x=0 it’s not differentiable since:

lim l0+hl –l0l h 0 h

= lim lhl . h 0 h

, which approaches to +1 if h 0 –1 if h0

Index FAQ

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

corner

f x x

cusp

2

3f x x

vertical tangent

3f x x

discontinuity

1, 0

1, 0

xf x

x

Differentiability

Index FAQ

If the derivative of a function is its slope, then for a constant function, the derivative must be zero. There is no change...

0dc

dx

example: 3y

0y

The derivative of a constant is zero.

Derivatives of some elementary functions

Index FAQ

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

This is part of a pattern.

1n ndx nx

dx

examples:

4f x x

34f x x

8y x

78y x

power rule

Derivatives of some elementary functions

Index FAQ

Find the horizontal tangents of:

4 22 2y x x 34 4

dyx 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.)

Rules for Differentiation

Index FAQ

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. )


Index FAQ

2

0

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

Derivatives of Trigonometric Functions

Index FAQ


h

xsin)hxsin(lim)'x(sin

0h

h

xsinxcoshsinhcosxsinlim

0h

h

xcoshsinlim

h

)1h(cosxsinlim

0h0h

h

xcoshsin)1h(cosxsinlim

0h

Proof

h

xh

h

hxx

dx

dhh

cossinlim

)1(cossinlimsin

00

Index FAQ

Derivative of the cosine Function

h

xcos)hxcos(lim)'x(cos

0h

h

xsinhsinlim

h

)1h(cosxcoslim

0h0h

h

xsinhsin)1h(cosxcoslim

0h

Find the derivative of cos x:

h

xsinhsinlim

h

)1h(cosxcoslim

0h0h

Index FAQ

Derivative of the cosine function is sine (cont.)

xsin1.xsin0.xcosh

hsinlimxsin

h

)1h(coslimosxc

h

xsinhsin

h

)1h(cosxcoslim

h

xsinhsin)1h(cosxcoslim

h

xcosxsinhsinhcosxcoslim

0h0h

0h

0h

0h

Index FAQ

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

tand

xdx

sin

cos

d x

dx x

2

cos cos sin sin

cos

x x x x

x

2 2

2

cos sin

cos

x x

x

2

1

cos x2sec x

2tan secd

x xdx


Index FAQ

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


Index FAQ

The Derivatives of the Sum, Difference, Product and Quotient

If and are derivable, and is any constant, u x v x C

then so is , , , and

. Its derivative is given by the formula

u x v x u x v x Cu x

u x

v x

(1) ( ) ( ) ( ) ( )u x v x u x v x

(2) ( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x u x v x

(3) ( ) ( )Cu x Cu x 2

( ) ( ) ( ) ( ) ( )(4) ( ( ) 0)

( ) ( )

u x u x v x u x v xv x

v x v x

Index FAQ

Proof

(1) Let ( ) ( ), we have to examiney u x v x

0 0

( ) ( ) ( ) ( )lim limx x

y u x x v x x u x v x

x x

0

( ) ( ) ( ) ( )limx

u x x u x v x x v x

x

(1) ( ) ( ) ( ) ( )u x v x u x v x

Index FAQ

0lim( ) ( ) ( )x

u vu x v x

x x

Thus ( ) ( ) is derivable and

( ) ( ) ( ) ( )

u x v x

u x v x u x v x

A similar argument applies to ( ) ( ),

that is

( ) ( ) ( ) ( )

u x v x

u x v x u x v x

(1) ( ) ( ) ( ) ( )u x v x u x v x

Index FAQ

x 0

(2) Let ( ) ( ), then we express in terms

of and . Finally, we determine by

examining lim

y u x v x y

u v y x

y

x

0 0

( ) ( ) ( ) ( )lim limx x

y u x x v x x u x v xy

x x

0

[ ( ) ][ ( ) ] ( ) ( )limx

u x u v x v u x v x

x

(2) ( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x u x v x

Index FAQ

0

( ) ( )limx

u x v v x u u v

x

0lim[ ( ) ( ) ]x

u v vv x u x ux x x

( ) ( ) ( ) ( )u x v x u x v x

Thus, ( ) ( ) is derivable and

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

u x v x

u x v x u x v x u x v x

Index FAQ

HOMEWORK!!

(3) ( ) ( )Cu x Cu x

Index FAQ

2

( ) ( ) ( ) ( ) ( )(4) ( ( ) 0)

( ) ( )


v x v x

0 0

( ) ( )( ) ( )

lim limx x

u x x u xy v x x v x

yx x

0

( ) ( )( ) ( )

limx

u x u u xv x v v x

x

Index FAQ

0 0

( ) ( )( ) ( )

lim limx x

u x x u xy v x x v x

yx x

0

( ) ( )( ) ( )

limx

u x u u xv x v v x

x

2

( ) ( ) ( ) ( ) ( )(4) ( ( ) 0)

( ) ( )


v x v x

Index FAQ

0

[ ( ) ] ( ) ( )[ ( ) ]lim

[ ( ) ] ( )x

u x u v x u x v x v

v x v v x x

0

( ) ( )lim

[ ( ) ] ( )x

uv x u x v

v x v v x x

0

( ) ( )lim

[ ( ) ] ( )x

u vv x u xx xv x v v x

Index FAQ

dy dy du

dx 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

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

Index FAQ

Remark

f(g(x))’= f ’(g(x)) g’(x) says that to get the

derivative of the “nested functions” you multiply

the derivative of each one starting from left to

right and so on

Index FAQ

Example : Find y’(1) for y = (3x2-2)3( 5x3-x-3)4

y ’= 3(3x2 -2)2 (3x2-2)’( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 ( 5x3-x-3)’

y ’= 3(3x2 -2)2 (6x) ( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 (15x2-1)

y ’(1) = 3(3-2)2 (6) (5-1-3)4 + (3-2)3 4 (5-1-3)3 (15-1)

YOUR TURN!, find when x=1 .

= 74

dy .

dx

. 2 x - 1 √5x2+4

For y =

Example for using Chain rule

Index FAQ

2sin 4y x

2 2cos 4 4d

y x xdx

2cos 4 2y x x


Index FAQ

2cos 3d

xdx

2cos 3

dx

dx

2 cos 3 cos 3d

x xdx

2cos 3 sin 3 3d

x 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”!)


Index FAQ

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 y

dx dx dx Do the same thing to both sides.

2 2 0dy

x ydx

Note use of chain rule.

2 2dyy xdx

2

2

dy x

dx y

dy x

dx y

Implicit Differentiation

Index FAQ

22 siny x y 22 sin

d d dy x y

dx dx dx

This can’t be solved for y.

2 2 cosdy dy

x ydx dx

2 cos 2dy dy

y xdx dx

22 cosdy

xydx

2

2 cos

dy x

dx y

This technique is called implicit differentiation.

1 Differentiate both sides w.r.t. x.2 Solve for y’


Index FAQ


Implicit Differentiation Process

1. Differentiate both sides of the equation with respect to x.

2. Collect the terms with y’=dy/dx on one side of the equation.

3. Factor out y’=dy/dx .

4. Solve for y’=dy/dx .

Index FAQ

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 0dydy

x yx ydxdx

Note product rule.

2 2 0dy dy

x x y ydx dx

22dy

y xy xdx

2

2

dy y x

dx y x

2 2 1

2 2 1m

2 2

4 1

4

5


Index FAQ

Find the equations of the lines tangent and normal to the

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

4

5m tangent:

42 1

5y x

4 42

5 5y x

4 14

5 5y x

normal:

52 1

4y x

5 52

4 4y x

5 3

4 4y x


Index FAQ

Find if .2

2

d y

dx3 22 3 7x y

3 22 3 7x y 26 6 0x y y

26 6y y x 26

6

xy

y

2x

yy

2

2

2y x x yy

y

2

2

2x xy y

y y

2 2

2

2x xy

y

x

yy

4

3

2x xy

y y

Substitute back into the equation.

y


Index FAQ

siny x

1siny xWe can use implicit differentiation to find:

1sind

xdx

1siny x

sin y xsin

d dy x

dx dx

cos 1dyydx

1

cos

dy

dx y

Derivatives of Inverse Trigonometric Functions

Index FAQ

We can use implicit differentiation to find:

1sind

xdx

1siny x

sin y xsin

d dy x

dx dx

cos 1dyydx

1

cos

dy

dx y

2 2sin cos 1y y

2 2cos 1 siny y 2cos 1 siny y

But2 2

y

so is positive.cos y

2cos 1 siny y

2

1

1 sin

dy

dx y

2

1

1

dy

dx x


Index FAQ

1siny x

1

cos

dy

dx y


)cos(sin

11 xdx

dy

21

1

xdx

dy

xy sin

1cos dx

dyy

Index FAQ


)(tansec

112 xdx

dy

21

1

xdx

dy

xy tan

1sec2 dx

dyy

Find xdx

d 1tan

xy 1tan

ydx

dy2sec

1

Index FAQ

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 e

h

definition of derivative

Derivatives of Exponential and Logarithmic Functions

Index FAQ

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 ee

dx h

0lim

x h x

h

e e e

h

0

1lim

hx

h

ee

h

0

1lim

hx

h

ee

h

1xe xe

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


Index FAQ

xe is its own derivative!

If we incorporate the chain rule: u ud due e

dx dx

We can now use this formula to find the derivative ofxa


Index FAQ

xda

dx

ln xade

dx

lnx ade

dx ln lnx a d

e x adx

Incorporating the chain rule:

lnu ud dua a a

dx dx


aaadx

d xx ln

Index FAQ

So far today we have:

u ud due e

dx dx lnu ud du

a a adx dx

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


Index FAQ

lny xye x

yd de x

dx dx

1y dyedx

1y

dy

dx e

1ln

dx

dx x

1ln

d duu

dx u dx


Index FAQ

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

logd

xdx

ln

ln10

d x

dx

1ln

ln10

dx

dx

1 1

ln10 x

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

1log

lna

d duu

dx u a dx


Index FAQ

u ud due e

dx dx lnu ud du

a a adx dx

1log

lna

d duu

dx u a dx

1ln

d duu

dx u dx


Index FAQ


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

xdx

dy

yln

11

xydx

dyln1

xxdx

dy x ln1

Documents

Index FAQ The derivative as the slope of the tangent line (at a point)