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The Chain Rule Quiz

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This quiz tests the work covered in the lecture on the chain rule and corresponds to Section 14.6 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There is a useful explanation and extra examples at http://www.math.hmc.edu/calculus/tutorials/multichainrule/. There are interactive exercises at http://www.math.temple.edu/ cow/ but you need to go to Book III, Functions, Chain Rule to find the questions. Pressing help gives you an explanation of how to use the chain rule. There are more web quizzes at Wiley, select Section 6. This quiz has 14 questions.


 

Question 1

 
 
Suppose z = 2xy where x = t2 +1 and y = 3 - t. Use the chain rule to determine which of the following is dz dt .
a) dz 2 dt = 6t- 4t - 2   b) dz = 4t3 + 6t- 6 dt
c) dz = 6t- 2 dt   d) dz = 12t- 6t2 - 2 dt

 

Not correct. Choice (a) is false.
Try again, you may have lost a t.
Not correct. Choice (b) is false.
Try again, check your partial derivatives.
Not correct. Choice (c) is false.
Try again, carefully watch your signs.
Your answer is correct.
∂z dx ---= 2y --= 2t ∂x dt
∂z dy ∂y-= 2x dt = - 1
dz dt = ∂z-dx- ∂z-dy ∂x dt + ∂y dt
  = 2y⋅2t- 2x
  = 2(3- t)⋅2t- 2(t2 + 1)
  =12t- 4t2 - 2t2 - 2 = 12t- 6t2 - 2 .
 

Question 2

 
 
Suppose z = xy- 2y2 where x = 3t+ 1 and y = 2t. Use the chain rule to determine which of the following is dz dt .
a) dz dt = - 4t+ 2   b) dz = - 16t- t dt
c) dz = 18t +2 dt   d) dz dt = - 10t2 + 8t+ 1

 

Your answer is correct.
∂z- dx- ∂x = y dt = 3
∂z-= x - 4y dy = 2 ∂y dt
dz dt = ∂zdx-+ ∂z-dy ∂x dt ∂y dt
  = y ⋅3+ (x- 4y)⋅2
  = 2x- 5y
  =2(3t+ 1)- 10t = - 4t+ 2.
Not correct. Choice (b) is false.
Try again, check your signs.
Not correct. Choice (c) is false.
Try again, check your signs.
Not correct. Choice (d) is false.
Try again, you may have not differentiated x and y with respect to t.
 

Question 3

 
 
Which of the following is the derivative, dz dt , written in terms of t where 2 2 z = x y+ y x where x = cost and y = t2 ?
a) dz= 2t2costsin t+ t4sint+ 2tcos2t+ 4t3cost dt   b) dz -- = - 2t2 costsint- 2t2sin t+ 4tcost+ 4t3 dt
c) dz dt = - 2t2costsin t- t4 sint+ 2tcos2t+ 4t3cost   d) dz = (2xy+ y2)(- sint)+ (x2 + 2xy)2t dt

 

Not correct. Choice (a) is false.
Try again, you may not have differentiated cos t correctly, check your signs.
Not correct. Choice (b) is false.
Try again, you may not have found the correct partial derivatives.
Your answer is correct.
-∂z 2 dx- ∂x = 2xy + y dt = - sint
∂z-= x2 + 2xy dy = 2t ∂y dt
dz dt = ∂z-dx+ ∂zdy ∂x dt ∂ydt
  = (2xy +y2)⋅sint+ (x2 + 2xy)⋅2t
  =(2t2cost+ t4)⋅(- sint)+ (cos2t+ 2t2cost⋅2t
  = - 2t2costsint- t4sin t+ 2tcos2 t+ 4t3cost.
Not correct. Choice (d) is false.
Try again, this is dz dt but not in terms of t.
 

Question 4

 
 
Suppose z = ln xy where x = √t and y = t2. Use the chain rule to determine which of the following is dz dt .
a) dz 1 2t3 dt = t2-+ √t-   b) dz 3 dt = 2t
c) dz 2 -- = - dt t   d) dz 1 -- = -- dt 2t

 

Not correct. Choice (a) is false.
Try again, you may have not have differentiated ln xy correctly.
Your answer is correct.
∂z 1 dx 1 ∂x = x- dt-= 2√t-
∂z 1 dy ∂y-= y- dt = 2t
dz dt = ∂z-dx- ∂zdy ∂x dt + ∂ydt
  = 1- -1-- 1- x ⋅2√t + y ⋅2t
  =√1-⋅-1√--+ 1-⋅2t t 2 t t2
  = 1-+ 1 = -3. 2t t 2t
Not correct. Choice (c) is false.
Try again, check your derivatives.
Not correct. Choice (d) is false.
Try again, carefully watch your signs when you differentiate.