Chain Rule - Kuta Software

Kuta Software - Infinite Calculus

Name___________________________________

Differentiation - Chain Rule

Date________________ Period____

Differentiate each function with respect to x. 1) y = ( x 3 + 3)

5

3) y = (−5 x 3 − 3)

2) y = (−3 x 5 + 1)

3

4) y = (5 x 2 + 3)

4

5) f ( x) =

4

−3 x 4 − 2

6) f ( x) =

7) f ( x) =

3

−2 x 4 + 5

8) y = (− x 4 − 3)

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3

−2 x 2 + 1

−2

Worksheet by Kuta Software LLC

9) y = (3 x 3 + 1)(−4 x 2 − 3) 4

10) y =

( x 3 + 4) 5 3x4 − 2

11) y = (( x + 5) − 1) 5

12) y = (5 x 3 − 3) 5

4

4

−4 x 5 − 3

Critical thinking question: 13) Give a function that requires three applications of the chain rule to differentiate. Then differentiate the function.

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Worksheet by Kuta Software LLC

Kuta Software - Infinite Calculus

Name___________________________________

Differentiation - Chain Rule

Date________________ Period____

Differentiate each function with respect to x. 5

1) y = ( x 3 + 3) dy 4 = 5( x 3 + 3) ⋅ 3 x 2 dx 4 = 15 x 2 ( x 3 + 3)

3) y = (−5 x 3 − 3)

2) y = (−3 x 5 + 1)

dy 2 = 3(−3 x 5 + 1) ⋅ −15 x 4 dx 2 = −45 x 4 (−3 x 5 + 1)

3

4) y = (5 x 2 + 3)

dy = 3(−5 x 3 − 3) 2 ⋅ −15 x 2 dx 2 = −45 x 2 (−5 x 3 − 3)

5) f ( x) =

4

3

7) f ( x) =

3

− 2)

− 1 f ' ( x) = (−2 x 2 + 1) 2 ⋅ −4 x 2 2x =−

3 4

(−2 x

−2 x 4 + 5

8) y = (− x 4 − 3) 2

3(−2 x + 5)

2 3

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2

+ 1)

1 2

−2

dy = −2(− x 4 − 3) −3 ⋅ −4 x 3 dx 8x3 = (− x 4 − 3) 3

− 1 f ' ( x) = (−2 x 4 + 5) 3 ⋅ −8 x 3 3 8x3 =− 4

−2 x 2 + 1 1

− 1 f ' ( x) = (−3 x 4 − 2) 4 ⋅ −12 x 3 4 3x3 =−

(−3 x

4

dy = 4(5 x 2 + 3) 3 ⋅ 10 x dx 3 = 40 x(5 x 2 + 3)

6) f ( x) =

−3 x 4 − 2

4

3

-1-

Worksheet by Kuta Software LLC

9) y = (3 x 3 + 1)(−4 x 2 − 3) 4 dy 3 4 = (3 x 3 + 1) ⋅ 4(−4 x 2 − 3) ⋅ −8 x + (−4 x 2 − 3) ⋅ 9 x 2 dx 3 = x(−4 x 2 − 3) (−132 x 3 − 32 − 27 x)

10) y =

( x 3 + 4) 5 3x4 − 2 4

5

dy (3 x 4 − 2) ⋅ 5( x 3 + 4) ⋅ 3 x 2 − ( x 3 + 4) ⋅ 12 x 3 = dx (3 x 4 − 2) 2 4 3 x 2 ( x 3 + 4) (11 x 4 − 10 − 16 x) = (3 x 4 − 2) 2 11) y = (( x + 5) − 1) 5

4

3 dy 5 4 = 4(( x + 5) − 1) ⋅ 5( x + 5) dx 3

= 20(( x + 5) − 1) ⋅ ( x + 5) 5

12) y = (5 x 3 − 3) 5

4

4

−4 x 5 − 3 3

1

− 1 dy = (5 x 3 − 3) 5 ⋅ (−4 x 5 − 3) 4 ⋅ −20 x 4 + (−4 x 5 − 3) 4 ⋅ 5(5 x 3 − 3) 4 ⋅ 15 x 2 dx 4 2 3 5 x (5 x − 3) 4 (−65 x 5 − 45 + 3 x 2 ) =

(−4 x

5

− 3)

3 4

Critical thinking question: 13) Give a function that requires three applications of the chain rule to differentiate. Then differentiate the function.

(

6

)

7

Many answers: Ex y = ((2 x + 1) + 2) + 3 6 6 5 dy 5 5 4 = 7 ((2 x + 1) + 2) + 3 ⋅ 6((2 x + 1) + 2) ⋅ 5(2 x + 1) ⋅ 2 dx

(

5

)

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