Introduction Of Derivatives Assessment Test

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Quizzes Created: 636 | Total Attempts: 801,881

Questions: 10 | Attempts: 156 | Updated: Mar 21, 2023

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Introduction Of Derivatives Assessment Test - Quiz

Take our intelligent assessment test to assess your knowledge of derivatives, which is a tool of calculus, the instantaneous rate of change of a function at a point in its domain.

Questions and Answers

1.

Find the derivative of (2+3x)(1-x) with respect to x.
- A.
  
  6x-1
- B.
  
  1-6x
- C.
  
  6
- D.
  
  -3
Correct Answer
B. 1-6x

Explanation
The given expression is a product of two functions, (2+3x) and (1-x). To find the derivative, we can use the product rule. The product rule states that the derivative of a product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. Applying this rule, we get (2+3x)(-1) + (1-x)(3). Simplifying this expression gives -2-3x+3-3x, which further simplifies to -6x+1. Hence, the correct answer is 1-6x.

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2.

Find the derivative of the function y=2x2(x-1) at the point x= -1.
- A.
  
  - 6
- B.
  
  14
- C.
  
  16
- D.
  
  18
Correct Answer
C. 16

Explanation
To find the derivative of the function y=2x^2(x-1), we can use the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by u'(x)v(x) + u(x)v'(x). Applying this rule, we find that the derivative of y=2x^2(x-1) is y'=2(2x)(x-1) + 2x^2(1) = 4x(x-1) + 2x^2 = 4x^2 - 4x + 2x^2 = 6x^2 - 4x. Evaluating this derivative at x=-1, we get y'(-1) = 6(-1)^2 - 4(-1) = 6 - 4 = 2. Therefore, the correct answer is 2.

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3.

If y= ( 1+x)2, find dy/dx.
- A.
  
  X-1
- B.
  
  2+2x
- C.
  
  1+2x
- D.
  
  2x-1
Correct Answer
B. 2+2x

Explanation
The given question asks for the derivative of y with respect to x, which can be found using the power rule for differentiation. According to the power rule, when differentiating a function of the form (a + b)^n, the derivative is n(a + b)^(n-1) times the derivative of (a + b). In this case, y = (1 + x)^2, so the derivative of y with respect to x is 2(1 + x)^(2-1) times the derivative of (1 + x), which simplifies to 2(1 + x). Therefore, the correct answer is 2 + 2x.

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4.

The derivative of (2x+1) (3x+1) is...
- A.
  
  12x+1
- B.
  
  6x+5
- C.
  
  6x+1
- D.
  
  12x+1
Correct Answer
D. 12x+1

Explanation
The given expression (2x+1) (3x+1) represents the product of two binomials. To find the derivative of this expression, we can use the product rule of differentiation. According to the product rule, the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. Applying this rule, we get the derivative of (2x+1) (3x+1) as (2)(3x+1) + (2x+1)(3) which simplifies to 6x+2 + 6x+3 = 12x+1.

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5.

Find the derivative of sinb/cosb.
- A.
  
  Sec2 b
- B.
  
  Tan b cosec b
- C.
  
  Cosec b sec b
- D.
  
  Cosec2 b
Correct Answer
A. Sec2 b

Explanation
The correct answer is "Sec2 b". The derivative of sinb/cosb can be found using the quotient rule. The derivative of sinb is cosb, and the derivative of cosb is -sinb. Applying the quotient rule, we get (cosb*cosb - (-sinb*sinb)) / (cosb)^2, which simplifies to cos^2(b) + sin^2(b) / cos^2(b). Since sin^2(b) + cos^2(b) = 1, the derivative becomes 1 / cos^2(b), which is equivalent to Sec^2(b). Therefore, the correct answer is Sec^2(b).

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6.

What is the derivative of t2 sin (3t - 5) with respect to the variable t?
- A.
  
  2t sin (3t - 5) - 3t2 cos (3t - 5)
- B.
  
  6t cos (3t - 5)
- C.
  
  2t sin (3t - 5) + 3t2 cos (3t - 5)
- D.
  
  2t sin (3t - 5) + t2 cos 3t
Correct Answer
C. 2t sin (3t - 5) + 3t2 cos (3t - 5)

Explanation
The given expression is a product of two functions: t^2 and sin(3t - 5). To find the derivative of this expression with respect to t, we can use the product rule. According to the product rule, the derivative of a product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. Applying the product rule to the given expression, we get 2t sin (3t - 5) + 3t^2 cos (3t - 5).

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7.

Find the derivatives with respect to x of the function √(2 – 3×2).
- A.
  
  -2×2 / √(2 – 3×2)
- B.
  
  -3x / √(2 – 3×2 )
- C.
  
  -2×2 / √(2 + 3×2)
- D.
  
  -3x / √(2 + 3×2)
Correct Answer
B. -3x / √(2 – 3×2 )

Explanation
The correct answer is -3x / √(2 – 3×2).

To find the derivative of the function √(2 – 3x^2), we can use the chain rule. The chain rule states that if we have a composite function, f(g(x)), then the derivative is given by f'(g(x)) * g'(x). In this case, the outer function is the square root and the inner function is 2 – 3x^2.

The derivative of the outer function, √u, is 1 / (2√u).

The derivative of the inner function, 2 – 3x^2, is -6x.

Using the chain rule, the derivative of the function √(2 – 3x^2) is (1 / (2√(2 – 3x^2))) * (-6x), which simplifies to -3x / √(2 – 3x^2).

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8.

Given the function f(x) = x to the 3rd power – 6x + 2, find the value of the first derivative at x = 2, f’(2).
- A.
  
  6
- B.
  
  3×2 – 5
- C.
  
  7
- D.
  
  9
Correct Answer
A. 6

Explanation
The given function is a polynomial function of degree 3. To find the value of the first derivative at x = 2, we need to differentiate the function. The derivative of x to the 3rd power is 3x^2, and the derivative of -6x is -6. The derivative of a constant term is 0. Therefore, the first derivative of f(x) is 3x^2 - 6. Plugging in x = 2 into the first derivative, we get f'(2) = 3(2)^2 - 6 = 12 - 6 = 6.

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9.

Find the partial derivatives with respect to x of the function: xy2 – 5y + 6.
- A.
  
  Y2 – 5
- B.
  
  Xy – 5y
- C.
  
  Y2
- D.
  
  2xy
Correct Answer
A. Y2 – 5

Explanation
The partial derivative of the function xy^2 - 5y + 6 with respect to x is given by taking the derivative of each term with respect to x while treating y as a constant. The derivative of xy^2 with respect to x is y^2, the derivative of -5y with respect to x is 0, and the derivative of 6 with respect to x is 0. Therefore, the partial derivative with respect to x is y^2 - 0 + 0 = y^2. Since the constant term -5 does not have an x term, it does not affect the derivative.

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10.

The derivative of ln (cos x) is...
- A.
  
  Sec x
- B.
  
  –sec x
- C.
  
  –tan x
- D.
  
  Tan x
Correct Answer
C. –tan x

Explanation
The derivative of ln (cos x) can be found using the chain rule. The derivative of ln(u) is 1/u multiplied by the derivative of u. In this case, u = cos x, so the derivative of ln (cos x) is 1/(cos x) multiplied by the derivative of cos x. The derivative of cos x is -sin x. Therefore, the derivative of ln (cos x) is -sin x divided by cos x, which simplifies to -tan x.

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