Math 142 Week-in-Review # 8 (Implicit Differentiation & Antiderivatives)

Brief Overview of Section 5.8:

• Implicit Derivative: Sometimes we wish to find the derivative of a function but we cannot easily solve for y in terms of x.

In this case, we can use implicit differentiation to find the derivative by following these steps:

– Take the derivative with respect to the independent variable ( usually x when findingdydx or t when finding

dydt or

dxdt ) of

both sides. When taking the derivative of y-terms multiply bydydx or

dydt . When taking the derivative of x-terms multiply

bydxdx = 1 or

dxdt .

– When findingdydx , move all terms with

dydx (y0) in them to the left-hand side and all the terms without

dydx (y0) in them to

the right-hand side.

– Factor outdydx (y0) from all terms on the left-hand side and solve for

dydx (y0).

• Related Rates: In a related rates problem, we wish to compute the rate of change of one quantity in terms of the rate of

change of another quantity. If we are not given an equation that relates the two quantities, then we need to first find such an

equation and then take the implicit derivative with respect to time.

1. Finddydx for each of the following.

(a)3

p(9+ y2)4 + ex = 2x4

.

(b) 4x5y2 +3x1/3y3 = lnx+ lny.

dy I dx -

- y'

rewrite -

mm

⇒ I 9 t y 23413 te"

=2×4

Eff" "

t I -

- I!? )

4.gl?wtyw2Y3.l2y.yMtex=8x3-4-zl9ty2Y?2y.y

'= 8×3 - ex

Fit ÷⇒"¥¥i:E

⇐- u - -.

.

( 20×4 y'

t 2g .4×5 y

' ) t ( x

-

43y3t3y2 . y'

.3×43 ) = If

thy . y'

-- ur

8×5 y y'

+ 9 y2×43 y

'

. ly y'

= Lx - 20×442 - x- 21343i¥E

.

' " " " "⇒ i=÷i÷÷::÷:÷:

18×54+9 y' x' 13

- Ly ) I8x5yt9y2x" 3- Ly )

Math 142 WIR, c�Maya Johnson, Spring 2019

(c) f (x) = log4(2xy)+ exy = 9

(d) 6xey +x+1

y+1= p2

2. Given xy2 + x2y = 0, find the equation of the line tangent to the curve at x = 3.

2

three logo,AB=logqAtleS4Blogo

,LzDtlog4yte=9wftp.yt

✓

FIT

q*+×e×Y)y'

= - ¥.

- Ye"E-+ ÷qoY' te

" '. l Yt ) -

-O

¥¥yy ¥¥e")Xx'M~+yexS+×y'e×"

--

O'

¥÷:÷÷÷i÷÷⇒i=÷÷÷:IIFait y

'. - two,

- yet

y

⇐w

- Titty . k¥477

I 62 ' they ' )ti'-

-

0 > ( 6×21 ,lyfgz)y'=-6eY_t6e*t6xeYy'tlYfp

- It't'

= o i¥) lbxe "- K¥77)

i.EE#iiE-iee⇒ i=¥×=-:÷÷÷÷

( 6xeY - ¥tp)Y' =- bei - ¥

Need Slope dYldxl×=z=Y4×=3e.

Find Y'

:

xuyjtxjyu-E.iq#fyqzx.2f*.y7--7y'=-y2-2xy.Need to find y

Zxytx when X =3 .

( y2t2xyy'

) t ( 2xytX2y ' ) -

- Oplugin × =3 into xy2tx2y=0 :

- -

Zxyy 't x' y'

=- y

'- Zxyy 342+1354--0=7342+94--0

÷÷÷÷÷÷÷÷÷÷÷÷f÷÷¥÷÷

.-

Tangent Line I 3,0 ) :

Slope at ( 3 ,-3 ) : slope m

-

- o :

Y' has ,=;¥4h÷ii¥÷E=-±

iii:÷i÷÷÷÷*÷÷¥÷÷⇐

o.mx . .

Math 142 WIR, c�Maya Johnson, Spring 2019

3. Suppose x(t) and y(t) are both functions of t. If y =p

2x2 +1

(a) finddydt

when x = 2, ifdxdt

= 5.

(b) finddxdt

when x = 2, ifdydt

= 7.

4. Suppose x(t), y(t), and z(t) are all functions of t. If x2 + y2 = z2,

dxdt

=�4,dzdt

= 5 finddydt

when x = 6, y = 4, and z = 3.

3

w

Et3 4=(2×2+1)" 2

dy Idt -- yl ;

dxldt -

- x'

?-

-

* y'

-

-

t.ly#tu5.64x.x

'

) ( x i x )

y 's 'zlze5"2. (4cs

)

y'

= Iz . Iz . 40=203 ⇒y'=dyz=20①

,-

-

From above : y' -12125415

" ? 4 x. x'

⇒ y' -12.4×12×2+15

" ? x'

zkkzczii.IT#-- IT ⇒x!d¥=

¥⇒⇒.

.

' ⇒ '"" '

⇒ y'

-

-

2×12×2+1542 x'

y'

w - u

,-

- -

2x . x 't 2 y y'

= Zz at'

-

Zyy'

=ZZZ'

- 2x x

'

Ty¥

y'

= ZZ'-y ( x ! - 4,

2-1=5 ,x=6 , y = 4

,2- =3 )

y'

= 137153-4611-41-314⇒ yh¥=3

Math 142 WIR, c�Maya Johnson, Spring 2019

5. Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant

rate of 2 m/s, how fast is the area of the spill increasing when the radius is 42 m?

6. Each side of a square is increasing at a rate of 5 cm/s. At what rate is the area of the square increasing when the area of the

square is 169 cm2?

7. If the demand equation of a certain commodity is given by x = 10�0.1p and the number of items manufactured (and sold)

is increasing at a rate of 20 per week, find the rate of change of revenue with respect to time when x = 4

4

r'

e 2 m Is

- -

- -

A '= ?

A -

- T' r"

* = ar'

A'

= Fur orl ( r

'=

2 's r =42 )

A ' 14,414=168a ⇒ A'=d¥=l6- - -

¥693 -

' Z

D. . FEI. . .

.

' A' " "

÷n¥÷÷÷÷÷÷*nu-

k I-

X = 10 - O .I p c- Solve for p .

.

X - 10 = - O . I p.

⇒=

.off' ¥

⇒ - fax t too =p ⇒ p = - 10 x t too

.

.

.

-

Revenue :

R -

- x . p = x ( - 10 x t too )

R =- 10×2 t 100 Xki:÷:;: .

"¥*.o

Math 142 WIR, c�Maya Johnson, Spring 2019

8. Two cars start moving from the same point. One travels south at 48 mi/h and the other travels west at 20 mi/h. At what rate

is the distance between the cars increasing three hours later?

9. A 10-foot ladder leans against a wall and slides down, with the foot of the ladder observed to be moving away from the

wall at a rate of 4 ft/s when it is 6 feet from the wall. At what rate is the top of the ladder moving downward?

5

y 's 48mi/h y'

= 20mi/h

- --

-

•×-•• Pythagorean Theorem e.

"

'

I'

,,↳.

448mi/h XZTYZ -

- E * *

? Find Z'

: X2ty2=z2

find × after 3 hours : 2××'

+ Zyy'

= ZZZ'

÷÷÷÷÷÷÷÷÷÷÷÷:÷÷⇒↳x¥⇒.¥¥i*⇒ Z'=dd#=52m

-

-

, µµx

" -4ft 's Fleet

×o£Use nifty 2=102 * * ( Pythagorean

Theorem )

Solve far ye .

42=102 - XZ-

⇒ y-

- ME = y-

Tobi⇒ y = 8 feet .

7 2 > y'

=

-2¥'

Find y'

: -ZyZy

x2tyj=§ yin .

- * = -16%44=-3zx.x.tk'"

⇒i;;:e÷a;Tmsdm÷dJ

Zyy'

=- 2x x

' Y' = - 3ft Is =.

Math 142 WIR, c�Maya Johnson, Spring 2019

Brief Overview of Section 6.1:

• Finding the General Antiderivative: If F 0(x) = f (x) , then the general antiderivative of f (x) is F(x)+C, where Cis an arbitrary constant.

– The collection of all antiderivatives of a function f (x) is called the indefinite integral, and is denoted byZf (x) dx (the indefinite integral of f (x) with respect to x). If we know one function F(x) for which F 0(x) = f (x),

then

Zf (x) dx = F(x)+C.

– Rules of Integration:Zk f (x) dx = k

Zf (x) dx (Constant Multiple Rule)

Z[ f (x)±g(x)] dx =

Zf (x) dx±

Zg(x) dx (Sum/Difference Rule)

Zxn dx =

1

n+1xn+1 +C,n 6=�1 (Power Rule)

Z1

xdx = ln |x|+C (Indefinite Integral of x�1 =

1

x)

Zex dx = ex +C (Indefinite Integral of Exponential Function)

• Finding a Specific Antiderivative: If F(x)+C is the general antiderivative of a function f (x), then finding a specific

antiderivative of f (x) involves finding a specific value for the constant C. In order to solve for C, we need to know a

point on the graph of f (x) such as (x1,y1) ( f (x1) = y1).

10. Find the most general antiderivative of the following functions. (Use C for the constant of integration. Remember to use

absolute values where appropriate.)

(a) f 0(x) = 4x3 �33x2 �12

(b)

Z ⇣�4

3p

x4 �2ex⌘

dx

(c)

Z ✓x5 �5x4 +6

x6

◆dx

6

-- - -

-

qY - -

-

*-

*-

*-

-

ft x ) =4414/33×337%2

× t C ⇒ fCx2=x4_ilx3-l2xt#

S I y" 3-

Eef) dx

= -4yg -

2e"

t C = -Eg×43_2e×t#

Erie Sf'In -5¥ t ¥ ) dx -

-

flip-

Exit7 dx

= In Ix I - 5 x- '

t 6s

t C =

lnlxlt-x-t-6qx-ST.ieI

Math 142 WIR, c�Maya Johnson, Spring 2019

(d) f 0(x) =10e�x �14

2e�x

(e)

Z ✓�3

x+

7

4x3� 2

x6

◆dx

(f)

Z �5� x3

��7� x7

�dx

11. Given that f 0(x) =p

x5 +3p

x7 +7ex, and f (0) = 22, find f (x).

7

rewrite 10 e- ×

=-

= 5 - Ze =

f- Cx ) = Ssw-

dx=5x-7e×t#

SfZ* 24×-3 - 2x-

6) dx -

- Sf 3. I + I,

x- Z 2x

-

6) dx

-

=- 31hL x I t If .

- 2¥'t

C-e-3lnlxt-154-zx-II.my#T

te

ft-

EE-

2×3two) dx

23-sx-5xgI-7xftx.IT#

f'

l x ) = ×

'

K+ × 2/3+7 exe

-

f- Lx ) -

- SHIK"t Let)dx = + x÷} t

7e× t C

= Ig ×2/2

+ go ×

' 013+ Tex to

Find C e.

f- l 07=22 #

f- to ) =3¥473¥17Ie t C = 22

7 t C = 22⇒ e=

- 7- 7

flxkzjktzox.IO/3tIexT#

Math 142 WIR, c�Maya Johnson, Spring 2019

12. The marginal profit from the sale of a certain product is given by MP(x) = 250ex �3x2, where x represents the number of

weeks since the product was made available for sale. Determine P(x), given that P(0) = 6000

13. The marginal cost function for a particular item is given by MC(x) = �3.50x+ 45 dollars per item. If the total cost for

producing 25 items is $1,500, find the fixed cost associated with producing this item.

8

derivative-

MP Cx ) = 250 ex - 3×2 -- p

¥oT

Pdx ) -- f(zsoef- 3)dx = 250 ex - 3¥ to = 250 ex - XI C

Find C g PCO ) = 6000

PCO ) = 2e°- lots t c = GoodPCx)=250e×-×3t575O

250 t C = Goo o

- 250- zgo

⇒ C = 5750

derivative-

= -

Solve for D Clo) ?

Mcl x ) =-

3.50×+45 = C' Cx )

( ( × ) = ) ( - 3. so × +45 )dx =-3.5-0×2+45 x t D

✓U 2

= - I .75×2+45 x t D

-

Find D : C (

253=1500C

125) = - I .75125 ) 't 45125 ) t D = 1500

⇒31 .

25 t D = 1500

- 31 .

-25- 31 .

25

⇒ D= 1468.75

Ctx ) =- I .

75×2+45 x t 1468.75

Find Fixed cost ( Clo ) )

Clo ) =- to -7510174510 ) t 1468.75

=$l468