### Multiple Choice Questions in Differential Calculus

1. What is the derivative of f(x) = x^3 at x = 2?

a) 6

b) 12

c) 24

d) 8

2. What is the second derivative of f(x) = sin(x)?

a) cos(x)

b) -sin(x)

c) -cos(x)

d) -cos(x)

3. What is the critical point of f(x) = x^2 - 4x + 5?

a) x = 2

b) x = 4

c) x = -2

d) There are no critical points

4. If f(x) = e^x + x^2, what is f'(x)?

a) e^x + 2x

b) e^x + x

c) 2e^x + x

d) e^x + 2

5. What is the limit of (x^3 - 3x^2 + 2x + 1) / (x^2 - x - 2) as x approaches 2?

a) 0

b) 1

c) 2

d) the limit does not exist

6. What is the derivative of f(x) = ln(x) + 2x?

a) 1/x + 2

b) 1/x + 2x

c) ln(x) + 2

d) ln(x) + 2x

7. What is the critical point of f(x) = x^3 - 12x + 5?

a) x = 2

b) x = -2

c) Both a and b

d) There are no critical points

8. If f(x) = 3x^4 + 2x^3 - 6x^2, what is f''(x)?

a) 24x^2 + 12x - 12

b) 12x^2 + 12x - 12

c) 12x^2 + 6x - 12

d) 24x^2 + 6x - 12

9. What is the derivative of f(x) = sqrt(x) / (x^2 + 1)?

a) (1/2)x^(-1/2)(x^2+1)^(-1) - 2x(x^2+1)^(-2)

b) (1/2)x^(-1/2)(x^2+1)^(-1) - x(x^2+1)^(-2)

c) (1/2)x^(-1/2)(x^2+1)^(-2) - x(x^2+1)^(-1)

d) (1 - 3 x^2)/(2 sqrt(x) (1 + x^2)^2)

10. What is the limit of (2x^2 - 3x + 1) / (x^2 - 1) as x approaches 1?

a) -2

b) -1

c) 1

d) 1/2

### Here are the solutions:

**1. b) 12 **

Using the power rule, f'(x) = 3x^2, so f'(2) = 12.)

**2. b) -sin(x)**

Taking the derivative of f(x) = sin(x) gives us f'(x) = cos(x). Taking the derivative again gives us f''(x) = -sin(x)

**3. a) x = 2 **

Taking the derivative of f(x) = x^2 - 4x + 5 gives us f'(x) = 2x - 4. Setting this equal to zero and solving for x, we get x = 2.

**4. a) e^x + 2x **

Using the sum rule and the derivative of e^x, we get f'(x) = e^x + 2x.

**5. d) the limit does not exist**

The limit of (x^3 - 3x^2 + 2x + 1)/(x^2 - x - 2) as x approaches 2 can be found by direct substitution: ((2)^3 - 3(2)^2 + 2(2) + 1)/((2)^2 - (2) - 2) = (8 - 12 + 4 + 1)/(4 - 2 - 2) = (1)/(0). Since division by zero is undefined, the limit does not exist

**6. a) 1/x + 2 **

Using the sum rule and the derivative of ln(x), we get f'(x) = 1/x + 2.

**7. c) Both a and b**

The derivative of f(x) = x^3 - 12x +5 is f’(x)=3x^2-12. To find the critical points we set f’(x)=0 and solve for x: 3x^2-12=0 => x^2-4=0 => (x+2)(x-2)=0 So we have two critical points: x=-2 and x=+2.

**8. (a)36x^2+12x-12**

If f(x)=3x4+2x3-6x^2 then the first derivative of f(x),f’(x)=12x3+6x2-12x and the second derivative of f(x),f’'(x)=36x^2+12x-12. The correct answer is (a)36x^2+12x-12.

**9. d) (1 - 3 x^2)/(2 sqrt(x) (1 + x^2)^2)**

Let's find the derivative of f(x) = sqrt(x) / (x^2 + 1) using the quotient rule and the chain rule:

f(x) = (x^(1/2)) / ((x^2 + 1))

f'(x) = [(1/2)(x^(-1/2))(x^2 + 1) - (x^(1/2))(2x)] / (x^2 + 1)^2

f'(x) = [x - 2x^(3/2)] / (2x^(3/2)(x^2 + 1)^2)

f'(x) = (x - 2x^2) / (2x^(5/2)(x^2 + 1)^2)

Simplifying this expression, we get:

f'(x) = (1 - 3x^2) / (2sqrt(x)(1 + x^2)^2)

**10. d) 1/2 **

If you try to plug in x=1 directly, you'll end up with an indeterminate form of 0/0. So, you need to use another method to evaluate this limit. One possible method is to factorize both the numerator and the denominator and simplify the expression. (2x^2 - 3x + 1) / (x^2 - 1) = [(2x - 1)(x - 1)] / [(x - 1)(x + 1)]. Therefore, the limit of (2x^2 - 3x + 1) / (x^2 - 1) as x approaches 1 is 1/2.

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