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Lesson: Higher-Order Derivatives Using the Chain Rule

Sample Question Videos

Worksheet • 15 Questions • 1 Video

Q1:

Suppose 𝑦 = 𝑧 8 and 𝑧 = βˆ’ π‘₯ βˆ’ 1 . Find d d 2 2 𝑦 π‘₯ .

  • A 5 6 ( βˆ’ π‘₯ βˆ’ 1 ) 6
  • B βˆ’ 8 ( βˆ’ π‘₯ βˆ’ 1 ) 7
  • C 5 6 ( βˆ’ π‘₯ βˆ’ 1 ) 8
  • D 4 2 ( βˆ’ π‘₯ βˆ’ 1 ) 6

Q2:

If 𝑦 = βˆ’ 6 𝑧 t a n and 𝑧 = βˆ’ 2 π‘₯ βˆ’ 3 , find d d 2 2 𝑦 π‘₯ .

  • A 4 8 ( 2 π‘₯ + 3 ) ( 2 π‘₯ + 3 ) t a n s e c 2
  • B 2 4 ( 2 π‘₯ + 3 ) ( 2 π‘₯ + 3 ) t a n s e c 2
  • C βˆ’ 2 4 ( 2 π‘₯ + 3 ) ( 2 π‘₯ + 3 ) t a n s e c 2
  • D 4 8 ( 2 π‘₯ + 3 ) ( 2 π‘₯ + 3 ) t a n s e c

Q3:

If 𝑦 = 2 𝑧 βˆ’ 1 3 𝑧 and 𝑧 = 2 π‘₯ + 5 2 , determine d d 2 2 𝑦 π‘₯ at π‘₯ = 0 .

  • A 4 7 5
  • B βˆ’ 8 3 7 5
  • C βˆ’ 2 1 5
  • D βˆ’ 1 1 5

Q4:

Given that 𝑦 = √ βˆ’ 𝑧 βˆ’ 8 and 𝑧 = βˆ’ 6 π‘₯ βˆ’ 3 2 , find d d 2 2 𝑦 π‘₯ at π‘₯ = βˆ’ 1 .

  • A βˆ’ 3 0
  • B βˆ’ 6
  • C9
  • D1

Q5:

Given that 𝑦 = ( 3 π‘₯ + 2 ) 4 , determine 𝑦 β€² β€² β€² .

  • A 6 4 8 ( 3 π‘₯ + 2 )
  • B 3 2 4 ( 3 π‘₯ + 2 )
  • C 1 2 ( 3 π‘₯ + 2 ) 3
  • D 2 4 ( 3 π‘₯ + 2 )
  • E 1 0 8 ( 3 π‘₯ + 2 ) 2

Q6:

Determine d d s i n 3 3 2 π‘₯ ο€Ί 7 π‘₯  .

  • A βˆ’ 1 3 7 2 1 4 π‘₯ s i n
  • B 1 3 7 2 1 4 π‘₯ s i n
  • C βˆ’ 2 8 7 π‘₯ s i n
  • D 9 8 1 4 π‘₯ c o s

Q7:

Suppose 𝑦 = 𝑧 3 and 𝑧 = βˆ’ π‘₯ + 5 . Find d d 2 2 𝑦 π‘₯ .

  • A 6 ( βˆ’ π‘₯ + 5 )
  • B βˆ’ 3 ( βˆ’ π‘₯ + 5 ) 2
  • C 6 ( βˆ’ π‘₯ + 5 ) 3
  • D 2 ( βˆ’ π‘₯ + 5 )

Q8:

Suppose 𝑦 = 𝑧 3 and 𝑧 = βˆ’ 5 π‘₯ βˆ’ 8 . Find d d 2 2 𝑦 π‘₯ .

  • A 1 5 0 ( βˆ’ 5 π‘₯ βˆ’ 8 )
  • B βˆ’ 1 5 ( βˆ’ 5 π‘₯ βˆ’ 8 ) 2
  • C 1 5 0 ( βˆ’ 5 π‘₯ βˆ’ 8 ) 3
  • D 5 0 ( βˆ’ 5 π‘₯ βˆ’ 8 )

Q9:

Suppose 𝑦 = 𝑧 7 and 𝑧 = βˆ’ π‘₯ + 4 . Find d d 2 2 𝑦 π‘₯ .

  • A 4 2 ( βˆ’ π‘₯ + 4 ) 5
  • B βˆ’ 7 ( βˆ’ π‘₯ + 4 ) 6
  • C 4 2 ( βˆ’ π‘₯ + 4 ) 7
  • D 3 0 ( βˆ’ π‘₯ + 4 ) 5

Q10:

Suppose 𝑦 = 𝑧 6 and 𝑧 = βˆ’ π‘₯ βˆ’ 6 . Find d d 2 2 𝑦 π‘₯ .

  • A 3 0 ( βˆ’ π‘₯ βˆ’ 6 ) 4
  • B βˆ’ 6 ( βˆ’ π‘₯ βˆ’ 6 ) 5
  • C 3 0 ( βˆ’ π‘₯ βˆ’ 6 ) 6
  • D 2 0 ( βˆ’ π‘₯ βˆ’ 6 ) 4

Q11:

If 𝑦 = βˆ’ 5 𝑧 t a n and 𝑧 = βˆ’ π‘₯ βˆ’ 7 , find d d 2 2 𝑦 π‘₯ .

  • A 1 0 ( π‘₯ + 7 ) ( π‘₯ + 7 ) t a n s e c 2
  • B 5 ( π‘₯ + 7 ) ( π‘₯ + 7 ) t a n s e c 2
  • C βˆ’ 1 0 ( π‘₯ + 7 ) ( π‘₯ + 7 ) t a n s e c 2
  • D 1 0 ( π‘₯ + 7 ) ( π‘₯ + 7 ) t a n s e c

Q12:

If 𝑦 = 𝑧 βˆ’ 1 𝑧 and 𝑧 = 2 π‘₯ + 1 2 , determine d d 2 2 𝑦 π‘₯ at π‘₯ = 0 .

  • A4
  • B βˆ’ 8
  • C βˆ’ 2
  • D βˆ’ 1

Q13:

If 𝑦 = 𝑧 βˆ’ 2 2 𝑧 and 𝑧 = 2 π‘₯ βˆ’ 3 2 , determine d d 2 2 𝑦 π‘₯ at π‘₯ = 0 .

  • A 4 9
  • B 8 2 7
  • C 2 3
  • D 1 3

Q14:

Given that 𝑦 = √ 8 𝑧 + 9 and 𝑧 = 6 π‘₯ βˆ’ 6 2 , find d d 2 2 𝑦 π‘₯ at π‘₯ = 1 .

  • A βˆ’ 2 0 8 3
  • B16
  • C 1 3 6 9
  • D 1 3

Q15:

Given that 𝑦 = √ 6 𝑧 + 1 and 𝑧 = βˆ’ 2 π‘₯ + 1 0 2 , find d d 2 2 𝑦 π‘₯ at π‘₯ = βˆ’ 1 .

  • A βˆ’ 7 3 2 3 4 3
  • B 1 2 7
  • C βˆ’ 5 9 4 3 4 3
  • D 1 7
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