Calculus Maximus WS 2.6: Chain Rule

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Name_________________________________________ Date________________________ Period______ Worksheet 2.6—The Chain Rule Short Answer Show all work, including rewriting the original problem in a more useful way. No calculator unless otherwise stated. 1. Find the derivative of the following functions with respect to the independent variable. (You do not need

to simplify your final answers here.)

(a) ( )32 7y x= − (b) 213 1

yt t

=+ −

(c) 21

3y

t⎛ ⎞= ⎜ ⎟−⎝ ⎠

(d) 3 3csc2xy ⎛ ⎞= ⎜ ⎟⎝ ⎠

(e) ( )23sec 1y tπ= − (f) 3 3sin siny x x= +

(g) 2 1tany xx

= (h) sec2 tan 2r θ θ= (i) ( ) 3 5csc 7f x =

They keykeymn

Rewrite RewriteorQuotientRule

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y2 31 1

Z

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Y µ T5 Y 372

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3

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Calculus Maximus WS 2.6: Chain Rule

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2. Find the equation of the tangent line (in Taylor Form) for each of the following at the indicated point.

(a) ( ) 2 2 8s t t t= + + at 2x = (b) ( ) 3 21

tf tt+=−

at ( )0, 2−

3. Determine the point(s) in the interval ( )0,2π at which the graph of ( ) 2cos sin 2f x x x= + has a

horizontal tangent. 4. Find the second derivative of each of the following functions. Remember to simplify early and often.

(a) ( ) ( )322 1f x x= − (b) ( ) ( )2sinf x x=

pointslopeformfNT

po.int2,4

ssiEITIEEI.amf'm.isE9EmIS'LH2tt2 x f f e 3 fi 7 3 2

12FEeg2s4429fH s zEdnu

equationtangentline y z if

tangentine Y 4 41 2 thatISThL2x

derivative 0 2S1nXc

FIX 2COSXt2S1nXCOSX 22sinx1 seniti 0f x 2scnxtzcoscxkcosx7 2scnxcs.mx Lf u

f2 2sinx1 0 SMX11 0

f x 25181 2052 2 SMH smx Inzy 251M I EIz

f X 2sinx 12C sin2x ZSirRX sink'z

F x 4sin2X 2Sinx 12 k Tf Cx 2ksindxtsinx t

f x 61 2 112127 flex cosCx4C2X productRole

fi II TIE7ffaa

tsmaaaxncuatcoscx.kz

Calculus Maximus WS 2.6: Chain Rule

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5. If ( ) ( )tan 2h x x= , evaluate ( )h x′′ at , 36π⎛ ⎞

⎜ ⎟⎝ ⎠. Simplify early and often.

6. If ( )5 3g = − , ( )5 6g′ = , ( )5 3h = , and ( )5 2h′ = − , find ( )5f ′ (if possible) for each of the following.

If it is not possible, state what additional information is required.

(a) ( ) ( )( )g x

f xh x

= (b) ( ) ( )( )f x g h x= (c) ( ) ( ) ( )f x g x h x=

(d) ( ) ( ) 3f x g x= ⎡ ⎤⎣ ⎦ (e) ( ) ( )( )f x g x h x= + (f) ( ) ( ) ( )( ) 2f x g x h x −

= +

ManfredderivativeunitcircleSidenoten'Cx sec42x7C27

Rewrite seditz

hi 2secC2X 2

Nyx 4csecc2 sed2HtmtanEHg.inf

H Ice 4fseccztfdllse.caeHtanC2tFdK2 rs

rsti'CEe4lsecEsIKsedEDltanCEDC274Cz Cz7Cr37C2 32f5J

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F 5 3gcs2g g

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ink.rsfisisi i iiiun'fF 5 27.6 notenoughinfof 57 162 togetgics

Calculus Maximus WS 2.6: Chain Rule

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Multiple Choice

_____ 14. If ( )2

1

3f x

x=

+, find ( )f x′ .

(A) ( )( )32 3

xf xx

′ = −+

(B) ( )2 3

xf xx

′ =+

(C) ( ) ( )2 3 2

xf xx x

′ = −+

(D) ( )( )32

1

2 3f x

x′ = −

+

(E) ( )2

233

x xf xx+′ = −+

_____15. If ( ) ( ) ( )31 4 1g x x x= − + , then ( )g x′ =

(A) ( )212 1 x− −

(B) ( ) ( )21 1 8x x− +

(C) ( ) ( )21 1 16x x− −

(D) ( ) ( )23 1 4 1x x− +

(E) ( ) ( )21 16 7x x− +

B f x 4 435t

f'Cx7 zCx43TEf 2xf'Cx 5

f x X_TEJ

CProducteldrain

Rule

gkxlf3CIXI.tn C4xtI 14 744

O 941 311 344 1 1C 24

947 1 74 314 1 1411 X

947 11 14 12 3 14 4 7

glad 1 51 16 1

Calculus Maximus WS 2.6: Chain Rule

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_____ 16. 52

23

5 9d xdx x

⎡ ⎤⎛ ⎞−⎢ ⎥ =⎜ ⎟⎜ ⎟⎢ ⎥−⎝ ⎠⎣ ⎦

(A) 2 4 2

2 610 ( 3) (10 17)

(5 9)x x x

x− −

−

(B) 2 4 2

2 510 ( 3) (5 16)

(5 9)x x x

x− − −

−

(C) 2 4

2 6240 ( 3)(5 9)x xx

− −−

(D) 2 4

2 660 ( 3)(5 9)x xx

−−

(E) 2 4

2 6100 ( 3)(5 9)x xx

−−

_____ 17. A derivative of a function ( )f x is obtained using the chain rule. The result is

( ) 33sec tanf x x x′ = . Which of the following could be ( )f x ?

I. ( ) 43 sec4

f x xπ= − +

II. ( ) 38 secf x x= +

III. ( ) 2sec sec tanf x x x x= +

(A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III

D adx 54siaY.E EI 695 37T

271529 1 2 3 Cox

fEyesNO

I FCK Tttseckx II fcxt8 sed.ae

fCx 8t eccxD3

f x _36eceakseoxtanx

iiiIi'sEi's c.mx e i isIoxunxfCX

3SeC4CX tanX ftp.secaatancxifqecxtanxtanxtlsecxktanxgseix