AP Calculus Unit 2 Progress Check Part B

ⅆ/ⅆx(tanx)=

sec^2x

ⅆ/ⅆx(cscx)=

−cscxcotx

Let g be the function given by g(x)=limh→0 sin(x+h)−sinx / h. What is the instantaneous rate of change of g with respect to x at x=π/3?

−√3/2. Correct. The function g is expressed as the limit of a difference quotient that represents the derivative of sinx. Therefore, g(x)=ddxsinx=cosx. The instantaneous rate of change of g with respect to x at x=π3 is g′(π3)=−sin(π3)=−3√2.

Below is an attempt to derive the derivative of secx using the product rule, where x is in the domain of secx. In which step, if any, does an error first appear? Step 1: secx⋅cosx=1 Step 2: ⅆⅆx(secx⋅cosx)=0 Step 3: ⅆⅆx(secx)⋅cosx−secx⋅sinx=0 Step 4: ⅆⅆx(secx)=secx⋅sinxcosx=secx⋅sinxcosx=secx⋅tanx

There is: no damn error

If g(x)=4cosx+2sinx+1, then g′(π/6)=

−2+√3 Correct. This question involves using the basic rules for the differentiation of trigonometric functions, and then evaluating the derivative at π6. g′(x)=−4sinx+2cosxg′(π6)=−4sin(π6)+2cos(π6)=−4⋅12+2⋅3√2=−2+3√

limh→0 5ex−5ex+h / 3h =

−5/3e^x Correct. limh→05ex−5ex+h3h=53limh→0ex−ex+hh=−53limh→0ex+h−exh=−53(ddxex)=−53ex

The function f is given by f(x)=(x^3+bx+6)g(x), where b is a constant and g is a differentiable function satisfying g(2)=3 and g′(2)=−1. For what value of b is f′(2)=0 ?

-22 Correct. Use the product rule to calculate f′(x), then evaluate at x=2 and set equal to 0, as follows. f′(x)=(3x2+b)g(x)+(x3+bx+6)g′(x)f′(2)=(12+b)(3)+(8+2b+6)(−1)=22+b=0⇒b=−22

If f(x)=√xcosx, then f′(x)=

cosx−2xsinx / 2√x Correct. The product rule is applied, as follows: f′(x)=ddx(x√)cosx+x√ddx(cosx)=12x√cosx+x√(−sinx)=cosx−2xsinx2x√.

Let g be the function given by g(x)=x^4−3x^3−x. What are all values of x such that g′(x)=12 ?

2.320 Correct. This question involves using the power rule for differentiation of polynomial functions to find g′(x)=4x3−9x2−1 and then solving an equation. The calculator is used to solve 4x3−9x2−1=12.

Let f be the function given by f(x)=x^3+3x^2−4. What is the value of f′(2) ?

24 Correct. This question involves using the power rule for differentiation of polynomial functions and then correctly evaluating at x=2. f′(x)=3x2+6x, so f′(2)=12+12=24.

If f(x)=4x^6−3x^4+2x^3+e^2, then f′(x)=

24x^5-12x^3+6x^2 Correct. This question involves using the power rule ddxxn=nxn−1 for differentiation of polynomial functions as follows. f′(x)=4⋅6x5−3⋅4x3+2⋅3x2+0

The graphs of the functions f and g are shown above. If h(x)=f(x)+4 / g(x)+2x, then h′(3)=

3/16 Correct. Use the quotient rule to calculate h′(x), then evaluate at x=3. h′(x)=f′(x)(g(x)+2x)−(f(x)+4)(g′(x)+2)(g(x)+2x)2h′(3)=f′(3)(g(3)+6)−(f(3)+4)(g′(3)+2)(g(3)+6)2The graph of f is used to determine that f(3)=4 and f′(3)=12. The graph of g is used to determine that g(3)=2 and g′(3)=−3. Therefore, h′(3)=12(2+6)−(4+4)(−3+2)(2+6)2=4+864=316.

If f(x)=2x^2−1 / 5x+3, then f′(−1)=

3/4 Correct. This question involves the use of the quotient rule to calculate the derivative of a function at a specific point. f′(x)=4x(5x+3)−5(2x2−1)(5x+3)2=10x2+12x+5(5x+3)2f′(−1)=10−12+5(−2)2=34

The table above gives the values of the differentiable functions f and g and their derivatives at x=4. What is the value of ⅆ/ⅆx(f(x)g(x)) at x=4 ?

31 Correct. The product rule is applied and correct values are chosen from the table, as follows. ⅆⅆx(f(x)g(x))∣∣x=4=f′(4)g(4)+f(4)g′(4)=7⋅3+2⋅5=31

What is the slope of the line tangent to the graph of y=9x^2 / x+2 at x=1 ?

5 Correct. The slope of the line tangent to the graph at x=1 is the value of the derivative of 9x2x+2 with respect to x at x=1. The quotient rule can be used to differentiate 9x2x+2, as follows. ⅆyⅆx=ⅆⅆx(9x2x+2)=ⅆⅆx(9x2)(x+2)−9x2ⅆⅆx(x+2)(x+2)2=18x(x+2)−9x2⋅1(x+2)2=9x2+36x(x+2)2ⅆyⅆx∣∣x=1=9x2+36x(x+2)2∣∣∣x=1=9⋅12+36⋅1(1+2)2=459=5