Find the derivatives of followings: 1. y = -6x1/6 ; w.r.t. x, 2. y = - sin (x) cot (x) ; w.r.t. x, 3. P = 4t ; w.r.t. t, 4. F = te-t ; w.r.t. t,

-x5/6, sin (x), 4t log t, e-t (1 - t)

-x-5/6, sin (x), 4t log t, e-t (1 + t)

-x-5/6, sin (x), 4t log t, e-t (1 - t)

-x-5/6, - sin (x), 4t log t, e-t (1 - t)

if y = xsin x, then find y',

x sin x [ (sin x)/x + ln x cos x]

x sin x [ (cos x)/x + ln x sin x]

x sin x [ (sin x)/x - ln x cos x]

x sin x [ (cos x)/x - ln x sin x]

Application Of Derivatives: Math Quiz! Test

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SS TIWARY

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Quizzes Created: 1 | Total Attempts: 250

Questions: 12 | Attempts: 250 | Updated: Mar 22, 2023

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Questions and Answers

1.

Find the derivatives of followings: 1. y = -6x^1/6; w.r.t. x, 2. y = - sin (x) cot (x) ; w.r.t. x, 3. P = 4^t; w.r.t. t, 4. F = te^-t; w.r.t. t,
- A.
  
  -x^5/6, sin (x), 4^t log t, e^-t (1 - t)
- B.
  
  -x^-5/6, sin (x), 4^t log t, e^-t (1 + t)
- C.
  
  -x^-5/6, sin (x), 4^t log t, e^-t (1 - t)
- D.
  
  -x^-5/6, - sin (x), 4^t log t, e^-t (1 - t)
Correct Answer
A. -x^5/6, sin (x), 4^t log t, e^-t (1 - t)

Explanation
The derivatives of the given functions are as follows:
1. The derivative of y = -6x^(1/6) with respect to x is -x^(-5/6).
2. The derivative of y = -sin(x) cot(x) with respect to x is -sin(x) - cos(x) cot(x).
3. The derivative of P = 4t with respect to t is 4.
4. The derivative of F = te^(-t) with respect to t is e^(-t) - te^(-t) (1 - t).

Therefore, the correct answer is -x^(-5/6), sin(x), 4t log(t), e^(-t) (1 - t).

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2.

If y = -sin (x), Find y''''
- A.
  
  Sin (x)
- B.
  
  - sin (x)
- C.
  
  - cos (x)
- D.
  
  Cos (x)
Correct Answer
B. - sin (x)

Explanation
The derivative of -sin(x) is -cos(x).

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3.

If y = x^x, find y' (note: if Xx is showing in answer, then it's x^x)
- A.
  
  X^x ln x
- B.
  
  X^x (1+ ln x)
- C.
  
  X^x / ln x
- D.
  
  X^x (1 - ln x)
Correct Answer
B. X^x (1+ ln x)

Explanation
The correct answer is xx (1+ ln x). This can be determined by applying the product rule of differentiation. The derivative of xx is 1, and the derivative of ln x is 1/x. Therefore, using the product rule, the derivative of xx (1+ ln x) is xx (1/x) + 1 (xx) = xx + xx ln x = xx (1+ ln x).

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4.

If y = x^{sin x}, then find y',
- A.
  
  X sin x [ (sin x)/x + ln x cos x]
- B.
  
  X sin x [ (cos x)/x + ln x sin x]
- C.
  
  X sin x [ (sin x)/x - ln x cos x]
- D.
  
  X sin x [ (cos x)/x - ln x sin x]
Correct Answer
A. X sin x [ (sin x)/x + ln x cos x]

Explanation
The given expression is y = xsin x. To find y', we need to differentiate y with respect to x. Using the product rule, the derivative of xsin x is (sin x) + x(cos x). Therefore, the correct answer is x sin x [ (sin x)/x + ln x cos x].

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5.

If f(x) = (1/4)x⁴ + (1/3)x³ -3x² + 20, then the number of critical points or points of extremum is:
- A.
  
  3
- B.
  
  4
- C.
  
  2
- D.
  
  Not defined
Correct Answer
A. 3

Explanation
The number of critical points or points of extremum can be determined by finding the derivative of the given function f(x) and then setting it equal to zero. In this case, the derivative of f(x) is f'(x) = x^3 + x^2 - 6x. By setting f'(x) = 0, we can solve for x and find the critical points. By solving the equation, we find that there are three critical points, which means the correct answer is 3.

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6.

Which of the following/s are not a critical point for the previous question?
- A.
  
  2
- B.
  
  -3
- C.
  
  0
- D.
  
  4
Correct Answer
D. 4
7.

If f(x) = x (2x-3)^1/2, the find the slope of the function at x =6,
- A.
  
  5
- B.
  
  6
- C.
  
  7
- D.
  
  8
Correct Answer
A. 5

Explanation
The slope of a function at a given point can be found by taking the derivative of the function and evaluating it at that point. In this case, the function f(x) = x(2x-3)^(1/2) can be rewritten as f(x) = x√(2x-3). Taking the derivative of this function with respect to x using the product rule and chain rule, we get f'(x) = √(2x-3) + x(2x-3)^(-1/2)(2). Evaluating this derivative at x = 6, we get f'(6) = √(2(6)-3) + 6(2(6)-3)^(-1/2)(2) = 5. Therefore, the slope of the function at x = 6 is 5.

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1
0
8.

Find the nature of extremum for the function, defined as, f(x) = x² - 3x + 2.
- A.
  
  Maximum at x = 2/3
- B.
  
  Maximum at x = -3/2
- C.
  
  Maximum at x = 3/2
- D.
  
  Minimum at x = 3/2
Correct Answer
D. Minimum at x = 3/2

Explanation
The given function is a quadratic function in the form of f(x) = x^2 - 3x + 2. To determine the nature of the extremum, we can find the derivative of the function, which is f'(x) = 2x - 3. Setting the derivative equal to zero and solving for x, we get x = 3/2.

Since the derivative is positive for x < 3/2 and negative for x > 3/2, we can conclude that the function has a minimum at x = 3/2. Therefore, the given answer "minimum at x = 3/2" is correct.

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9.

The function f(x) = 2x² - 8x + 4 is increasing in the interval from:
- A.
  
  (- ∞, 0]
- B.
  
  (- ∞, 2]
- C.
  
  [2 , ∞]
- D.
  
  None of these
Correct Answer
C. [2 , ∞]

Explanation
The function f(x) = 2x^2 - 8x + 4 is a quadratic function with a positive leading coefficient (2), which means it opens upwards. When a quadratic function opens upwards, it is increasing on the interval where its derivative is positive. To find the derivative of f(x), we can differentiate the function with respect to x. The derivative of f(x) = 2x^2 - 8x + 4 is f'(x) = 4x - 8. Setting f'(x) > 0 and solving for x, we get x > 2. Therefore, the function f(x) is increasing in the interval [2, ∞].

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10.

If a particle is moving with a variable velocity, as v = 2t² - 4t + 2, find the minimum attained velocity by the particle.
- A.
  
  2 m/s
- B.
  
  -2 m/s
- C.
  
  1 m/s
- D.
  
  0 m/s
Correct Answer
D. 0 m/s

Explanation
The minimum attained velocity by the particle can be found by taking the derivative of the velocity function with respect to time and setting it equal to zero. By differentiating v = 2t^2 - 4t + 2, we get dv/dt = 4t - 4. Setting this equal to zero, we find t = 1. Substituting this value back into the velocity function, we get v = 2(1)^2 - 4(1) + 2 = 0 m/s. Therefore, the minimum attained velocity by the particle is 0 m/s.

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11.

Two runners are running on a track, as, s = t(t-2), find the time after the starting point when they are crossing each other.
- A.
  
  2 sec
- B.
  
  1 sec
- C.
  
  3 sec
- D.
  
  None of these
Correct Answer
A. 2 sec

Explanation
The given equation represents the positions of the two runners on the track. By setting the equation equal to zero and solving for t, we can find the time when they are crossing each other. In this case, when t = 2, the equation becomes 0 = 2(2-2), which simplifies to 0 = 0. This means that both runners are at the same position at t = 2 seconds, indicating that they are crossing each other at that time. Therefore, the correct answer is 2 sec.

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12.

In the previous question, find the time at which they are running at the same speed:
- A.
  
  1 sec
- B.
  
  2 sec
- C.
  
  3 sec
- D.
  
  None of these
Correct Answer
A. 1 sec

Explanation
The answer is 1 sec because it is the only option given that indicates a time at which they are running at the same speed. The other options do not provide any information about the speed of the individuals at a specific time. Therefore, the correct answer is 1 sec.

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Current Version
Mar 22, 2023

Quiz Edited by
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May 18, 2020

Quiz Created by
SS TIWARY

Application Of Derivatives: Math Quiz! Test

Find the derivatives of followings: 1. y = -6x1/6 ; w.r.t. x, 2. y = - sin (x) cot (x) ; w.r.t. x, 3. P = 4t ; w.r.t. t, 4. F = te-t ; w.r.t. t,

If y = -sin (x), Find y''''

If y = xx, find y' (note: if Xx is showing in answer, then it's xx)

If y = xsin x, then find y',

If f(x) = (1/4)x4 + (1/3)x3 -3x2 + 20, then the number of critical points or points of extremum is:

Which of the following/s are not a critical point for the previous question?

If f(x) = x (2x-3)1/2, the find the slope of the function at x =6,

Find the nature of extremum for the function, defined as, f(x) = x2 - 3x + 2.

The function f(x) = 2x2 - 8x + 4 is increasing in the interval from:

If a particle is moving with a variable velocity, as v = 2t2 - 4t + 2, find the minimum attained velocity by the particle.

Two runners are running on a track, as, s = t(t-2), find the time after the starting point when they are crossing each other.

In the previous question, find the time at which they are running at the same speed:

Find the derivatives of followings: 1. y = -6x^1/6; w.r.t. x, 2. y = - sin (x) cot (x) ; w.r.t. x, 3. P = 4^t; w.r.t. t, 4. F = te^-t; w.r.t. t,

If y = x^x, find y' (note: if Xx is showing in answer, then it's x^x)

If y = x^{sin x}, then find y',

If f(x) = (1/4)x⁴ + (1/3)x³ -3x² + 20, then the number of critical points or points of extremum is:

If f(x) = x (2x-3)^1/2, the find the slope of the function at x =6,

Find the nature of extremum for the function, defined as, f(x) = x² - 3x + 2.

The function f(x) = 2x² - 8x + 4 is increasing in the interval from:

If a particle is moving with a variable velocity, as v = 2t² - 4t + 2, find the minimum attained velocity by the particle.