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Differentiating Powers and Polynomials Quiz

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This quiz tests the work covered in Lecture 12 and corresponds to Section 3.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There are more web quizzes again at Wiley, select Section 1. The function referred to in Question 13 is the function in Question 12.

There is a film at http://www.calculus-help.com/funstuff/tutorials/derivatives/deriv03.html which covers this topic well.

There are quizzes and web resources at http://quiz.econ.usyd.edu.au/mathquiz/differentiation/index.php.


 

Question 1

 
 
Which of the following statements are correct? There may be more than one correct answer.
a) If 2 f(x) = x + 2x +4 then ′ f (x) = 2x+ 2 .
b) If 3 2 y = x + 3x + 5 then -dy= 3x2 + 6x+ 5. dx
c) If f(x) = x-3 then f′(x) = - 3x -2.
d) If 1 y = √-- t then dy - 1 -dt =-√--3 . 2 t
e) If 2 1 h(z) = - z2 + z then ′ 1 h (z) = - z + 1.

 

There is at least one mistake.
For example, choice (a) should be true.
There is at least one mistake.
For example, choice (b) should be false.
The derivative of a constant is zero.
There is at least one mistake.
For example, choice (c) should be false.
- 3- 1 = - 4 so f′(x) = - 3x-4.
There is at least one mistake.
For example, choice (d) should be true.
Since 1 dy 1 3 - 1 y = t- 2 ⇒ dt = -2 t- 2 =-√--3 . 2 t
There is at least one mistake.
For example, choice (e) should be false.
Recall that h(z) = - 2z-2 + z-1 .
′ -4 1- h (z) = z3 - z2 .
Your answers are correct
  1. True.
  2. False. The derivative of a constant is zero.
  3. False. - 3- 1 = - 4 so f′(x) = - 3x-4.
  4. True. Since 1 dy 1 3 - 1 y = t- 2 ⇒ dt = -2 t- 2 =-√--3 . 2 t
  5. False. Recall that h(z) = - 2z-2 + z-1 .
    ′ -4 1- h (z) = z3 - z2 .
 

Question 2

 
 
Let y = 4s3 + 3s+ 7. Which one of the following statements is correct?
a) dy -ds = 8s +3 .   b) dy 2 ds = 12s + 10.
c) dy = 12s2 + 3. ds   d) None of the above, you cannot differentiate with respect to s.

 

Not correct. Choice (a) is false.
Try again, you have differentiated 2 y = 4s + 3s+ 7.
Not correct. Choice (b) is false.
Try again, recall that the derivative of a constant is 0.
Your answer is correct.
dy = 4× 3x2 + 3 = 12x2 + 3. ds
Not correct. Choice (d) is false.
Try again, the function can be differentiated, it doesn’t matter what the variable is called.
 

Question 3

 
 
A particle is moving in a straight line and the distance, in metres, from its starting position at time t , in seconds, is given by s(t) = 27t+ 7t2 for the first two seconds of its motion. Which of the following gives the formula for the velocity, v, of the particle, in metres per second for the first two seconds?
a) v(t) = 27+ 14t.   b) v(t) = 27+ 7t.
c) v(t) = 34.   d) There is not enough information available.

 

Your answer is correct.
v(t) = s′(t) = 27+ 7t2.
Not correct. Choice (b) is false.
Try again, v(t) = s′(t).
Not correct. Choice (c) is false.
Try again, this is the average velocity for the first two seconds.
Not correct. Choice (d) is false.
Try again, recall v(t) = s′(t).
 

Question 4

 
 
A particle’s velocity at time t is given by the formula v(t) = u+ at.
Which of the following could be the function describing the distance of the particle from its starting position.
a) s(t) = ut +at2.   b) s(t) = ut + 12at2 .
c) s(t) = u + 2at2.   d) None of the above.

 

Not correct. Choice (a) is false.
Try differentiating the function and see if it gives you v(t).
Your answer is correct.
s′(t) = u+ at = v(t) so this is the correct function.
Not correct. Choice (c) is false.
Try differentiating the function and see if it gives you v(t).
Not correct. Choice (d) is false.
Try differentiating the functions and see which one gives you v(t).