Calculus Chapter 2 Derivatives Section 2.1 Derivatives by the Formal Definition of the Derivative Day 1 What is a derivative?

• The slope of a tangent line. • The tangent line is a picture of the slope. • The purpose of knowing the slope of the tangent line is that it tells rate of change (speed!!).

How do we know we need to find a derivative.

• Asked to find the equation tangent line at a point. • Asked to find f’(x) given f(x) or f”(x) given f’(x) • Asked to find

€

ddx

or

€

dydx

or

€

ΔyΔx

• Asked to find velocity given displacement • Asked to find acceleration given velocity • Asked to find rate of change

From Slope to the Definition of the Derivative Through Limits Consider the slope between the 2 points shown to the right: Rewrite the slope using function notation Rewrite the slope in terms of

€

Δx Apply a limit to make the approximation of the limit into the exact slope:

This is called the Formal Definition of the Derivative of a Function Every Derivative we learn this year was created using this formula!

Section 2.1 Derivatives by the Formal Definition of the Derivative Day 2 When is a function NOT differentiable? Any time the slope to the left and the right of a point do not match.

Examples: 1. Discontinuities 2. Corners 3. When the slope is vertical (jump, hole, v. asymptote)

The relationship between differentiability and Continuity: 1. If a function is differentiable at a point, x = c, then it is continuous at x = c. Differentiability implies continuity. 2. It is possible for a function to be continuous at a point, x = c, and NOT be differentiable at x = c. Continuity does NOT imply differentiability. See examples 2 and 3 above.

Section 2.2 Basic Derivative Rules Day 1 Constants

• Constant functions never change, so their rate of change is 0. • The derivative of a constant function is 0. • Example: f(x) = 4 f’(x) = 0

Power Rule (for polynomials)

• For each term, multiply the power by the coefficient in front and decrease the power by 1. • Formal definition: if n is a rational number, then the function f(x) = xn is differentiable and is

equal to:

€

ddx[axn ] = anxn−1

• Example: f(x) = 3x3 – 4x2 + x – 6 f’(x) = 3•3x2 – 4•2x1 + 1x0 – 0 ! f’(x) = 9x2 – 8x + 1

• Examples: y = 2x5 – x4 + x – 1 dy/dx =

y = x7 + 3x2 + 8x + 2 d/dx = Interesting Uses of the Power Rule: find y’

€

y = x

y = x 43

y =2x 3

Derivative of Sine and Cosine • y = sin x y’ = cos x • y = cos x y’ = -sinx

Equation of the Tangent Line at a Point

• Step 1: Find the derivative - The slope of the tangent line is found using the derivative. • Step 2: plug the x value into the derivative - The slope is different along the curve of f(x),

plug the x value into f’(x) is find the slope at the point. • Step 3: state the equation in point-slope form

€

y − y1 = m(x − x1) at (x1,y1) Graphing a Derivative on the Calculator Step 1: put the original into Y1 Step 2: set Y2 = nDeriv(Y1) nDeriv( is the 8th option in the Math Menu Y1 can be found by pressing VARS, Right to Y-VARS, 1 for Function, 1 for Y1

Note: Your may want to leave this command in Y7 and turn it on/off as needed. Examples: j(t) = 4x3 – 2x + 6 Graph j(t) and j’(t) and use it to state the equation of the tangent line at the following values: x = -2 x = 1 x = 3

Displacement

• The distance (y) from our starting value over time (x) • One direction is positive, one direction is negative. • Displacement is the original equation f(x)

Picture: f(x) = 2x3 – 4x2 + 1

Velocity • The rate (dy/dx) at which displacement is changing over

time (x) • Moving in the positive direction results in a positive velocity. • Moving in the negative direction results in a negative

velocity (moving backwards). • Velocity is the derivative of f(x), written f’(x) • Speed is the absolute value of velocity (never negative).

Acceleration • The rate (dy2/d2x) at which velocity is changing over time (x) • Speeding up results in a positive acceleration. • Slowing down results in a negative acceleration. • Acceleration is the derivative of f’(x), written f”(x)

Calculus' Chapter'2' Derivatives'

Section'2.3'Rules'of'Derivatives'Product'Rule'' ' ' ' ' ' ' ' ' Day'1'

• The$Product$Rule$is$used$when$two$polynomials$are$being$multiplied$by$one$another.$$

• Formal$Definition:$The$derivative$of$(f$·$g)$is$the$derivative$of$the$first$function$times$the$second$

function,$plus$the$first$function$times$the$derivative$of$the$second$function.$$

[f(x)$g(x)]’$=$f’(x)$g(x)$+$f(x)$g’(x)$

$

• Example:$h(x)$=$(2x$+$3)$∙$(x2$–$4x$+$1)$ $ f(x)$=$ $ $ f’(x)$=$

$ $ $ $ $ g(x)$=$ $ $ g’(x)$=$

$$$$$

$$$$h’(x)$=$(2)$∙$(x2$–$4x$+$1)$+$(2x+3)$∙$(2x$–$4)$

$$$$h’(x)$=$2x2$–$8x$+$2$+$4x2$O$8x$+$6x$O12$

$$$$h’(x)$=$6x2$+$10x$+$10$

'

Quotient'Rule' ' ' Section'2=3' ' ' ' ' ' Day'2'

• The$Quotient$Rule$is$used$when$one$polynomial$is$being$divided$by$another.$

• Formal$Definition:$The$derivative$of$(f/g)$is$the$derivative$of$the$first$function$times$the$second$

function,$minus$the$first$function$times$the$derivative$of$the$second$function,$all$divided$by$the$

second$function$squared,$given$the$second$function$is$$$≠$0.$'

[f(x)/g(x)]’$=$f’(x)$g(x)$−$f(x)$g’(x)$'

$$$$$[g(x)]2 **NOTICE$THE$TOP$IS$THE$PRODUCT$RULE$WITH$A$SUBTRACTION$SIGN,$INSTEAD$OF$AN$ADDITION$SIGN**$

• Example:$h(x)$=$2x²$+$3x$O$4$$$ $ f(x)$=$ $ $ f’(x)$=$ $$$$$$$$$3x$ $ $ $ g(x)$=$ $ $ g’(x)$=

$$$ $$$$h’(x)$=$(4x$+$3)$(3x)$–$[(2x²$+$3x$–$4)$(3)]$$

[3x]2 $$$$h’(x)$=$12x²$+9x$–$[6x²$+9x$–$12]$'

$$$$$$$[3x]2 $$$$h’(x)$=$6x²$+$12$'

$$$$$$$$[3x]2 '

Rewriting'Functions' ' ' ' Section'2=3'' ' ' Day'3'

When$is$it$appropriate$to$rewrite$a$function?$

• When$you$have$a$variable$in$the$denominator.$

• When$you$are$dividing$by$a$single$integer.$

• When$you$can$factor$out$the$denominator$in$the$numerator.$$

• Anytime$it$would$be$easier$to$alter$the$problem$and$use$the$Power$Rule.$

$

$

Derivatives'of'Trig'Functions' ' ' Section'2=3'' ' ' Day'4'

• You$can$find$the$derivative$of$any$trig$function$if$you$know$your$trig$identities.$$

• For$instance,$tan$(x)$=$(sinx)/(cosx)$

• Therefore,$using$the$quotient$rule,$we$get:$

h’(tanx)$=$(cosx)(cosx)$–$[(sinx)(Osinx)]'

$$$$$$$$$$cos2x h’(tanx)$=$cos2x–$sin2x

$$$$$$$$$$$$cos2x h’(tanx)$=$$$$$$$$1$$$$$_$$$$$$

$$$$$$$$$cos2x $ $ h’(tanx)$=$sec²x

The'Chain'Rule' ' ' Section'2=4'' ' ' ' Day'1'

• The$Chain$Rule$is$used$when$you$have$to$find$the$derivative$of$a$function$f(g(x)).$

• In$words:$find$the$derivatives$by$working$outsideOin.$$Start$by$finding$the$outside$function$while$

you$leave$the$inside$function$alone.$$Then,$on$the$outside,$multiply$by$the$derivative$of$the$

inside$function.$

• Formal$Definition:$

[f(g(x))]’$=$f’(g(x))$∙$g’(x)$

Examples:$ y$=$(5x$–$7)3$ $ $

y = !! + !! − 1

! = !!!!

!

Day 2 Examples

! = !!!!!

! = !(!!!!)!

! = (! − 3)!(!! − !)!

y = cos(4x + 1

y$=$¼$sin2(2x)$

Calculus' Chapter'2'' Derivatives' ' ' Section'2.5'Implicit'Differentiation'

Implicit$&$Explicit$Functions:$

• Explicit$Form:$! = !! = !!!$

• Implicit$Form:$!" = 1$

When & How to use Implicit Differentiation:

• !!! !

! = 3!!$$!$The$variables$agree,$so$you$use$the$Power$Rule.$

• !!! !

! = 3!! !"!"$$!$The$variables$disagree,$so$you$use$the$Chain$Rule.$

• $

Guidelines for Implicit Differentiation

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

2. Collect all terms involving !"!" on the left side of the equation and move all other terms to the

right side of the equation.

3. Factor !"!" out of the left side of the equation.

4. Solve for !"!" by dividing both sides of the equation by the left-hand factor that does not contain !"!".

Examples

• !!" ! + 3! = !

!" ! + !!" 3! = 1+ 3 !"!"$$!$Distribute$the$

!!"$to$x$&$3y&,$then$follow$the$previous$

rules.$In$the$first$part,$the$variables$agree,$so$use$the$Power$Rule.$In$the$second$part,$the$

variables$disagree,$so$you$must$use$the$Chain$Rule.$$

$

$

• Find$d/dx$of$$$$xy2$=$3x$

Section'2.6'Related'Rates'Finding$Related$Rates:$

• We$now$use$the$Chain$Rule$to$find$the$rates$of$change$of$two$or$more$related$variables$that$are$

changing$with$respect$to$TIME.$

Part 1: non-calculus work.

Draw a half full soup can:

Picture a soup can that has dimensions h = 10 cm and r = 3 cm. We could find the volume of the can using:

!!!!!!!!!!! = !!!ℎ

Now Picture a machine pouring soup into the can at a rate of 100 cm3 / s. Find how long it takes to fill the

can.

As the can is being filled, is the radius of the soup within the can changing? Is the rate constant?

Part 2: basic calculus work.

As the can is being filled, is the height of the soup within the can changing? Is the rate constant?

Find the derivative of the volume formula with respect to time.

Find the rate of change of the height when the height of the soup in the can is 4 cm.

Part'3:'related'rates!'

Now(let’s(get(weird(M(M(Picture(a(soup(can((h(=(10(cm(and(r(=(3(cm)(that(is(growing(taller(and(taller(while(the(

radius(stays(the(same.((If(the(height(changes(at(2(cm/s(then(how(fast(is(the(volume(changing(at(the(instant(that(

the(height(reaches(20(cm?(

(

(

(

(

(

(

(

(

Not(weird(enough?((M((Fine((M(lets(say(the(soup(can(from(the(example(above(must(maintain(its(original(volume(as(the(height(is(growing(at(2(cm/s.((What(must(happen(to(the(radius?(((

(

(

What(is(the(radius(when(the(height(reaches(14(cm?(((

(

(

(

(

How(fast(will(the(radius(be(changing(when(the(height(reaches(14(cm?(

( (

Example(#5:(The(radius(of(a(circle(is(increasing(at(a(rate(of(3cm/min.((

(( (a)((Find(the(Rate(of(Change(of(the(area(when(r=6cm(

(

(

(

(

(

(b) Find(the(Rate(of(Change(of(the(circumference(when(r=24cm((

(

(

(

(

(

Example(#19:(A(spherical(balloon(is(inflated(with(gas(at(the(rate(of(800(cubic(cm/min.(((

(a)(Find(the(Rate(of(Change(of(the(volume(when(r(=(3(cm(((

(

(

(

(

(

(b)(Find(the(Rate(of(Change(of(the(surface(area(when(r(=(6(cm(