Differentiation of Composite & Inverse Functions (College Board AP® Calculus AB): Exam Questions

1 hour37 questions
1
1 point

If y=5 sin(x5), then d2ydx2=

  • −125sin(x5)

  • 15sin(x5)

  • −sin(x5)

  • −15sin(x5)

2
1 point

If f(x)=ex, then f'(x)=

  • xe(x−1)

  • ex2x

  • exx

  • ex

3
1 point

What is the slope of the line tangent to the curve y=arctan (2x) at the point at which x=12?

  • 1

  • 12

  • π4

  • 23

4
1 point

Let f be a differentiable function with an inverse function f−1.

If f(3)=5 and f'(3)=4 what is (f−1)'(5)?

  • 13

  • 14

  • 4

  • 3

5
1 point

If y=(4x3+5x+2)5, then dydx=

  • 5(4x3+5x+2)4

  • (12x2+5)5

  • 5(12x2+5)(4x3+5x+2)4

  • (12x2+5)(4x3+5x+2)4

1
1 point

If y=(x4+sinx)5, then y'=

  • 5(x4+sinx)4

  • 5(4x3+cosx)4

  • 5(4x3+cosx)4·(12x2−sinx)

  • 5(x4+sinx)4·(4x3+cosx)

2
1 point

Let f(x)=(3x−1)3 and let g be the inverse function of f. Given that f(0)=−1, what is the value of g'(−1)?

  • −14

  • 116

  • 19

  • 9

3
1 point

If f(x)=cos (e−2x), then f'(x)=

  • −sin(e−2x)

  • −sin(e−2x)−2e−2x

  • −2e−2x sin(e−2x)

  • 2e−2x sin(e−2x)

4
1 point

If f(x)=tan(2x), then f'(π2)=

  • 2

  • 0

  • −2

  • 1

5
1 point

A function f is defined by f(x)=arccos(2x).

What is the second derivative of f(x)?

  • −8x(1−4x2)3

  • −21−4x2

  • 8x(1−4x2)3

  • 21−4x2

6
1 point

Let f be the function defined by f(x)=4x2+2x for x>−14. If g(x)=f−1(x) and g(20)=2, what is the value of g'(20)?

  • 1162

  • 118

  • 18

  • 162

1
1 point

If f(x)=x2−13 and g(x)=4x+1, then the derivative of f(g(x)) at x=1 is

  • 103

  • 203

  • 363

  • 205

2
1 point

ddx(eex)=

  • eex

  • xeex

  • eex+x

  • eex2

3
1 point

If f(x)=x3+3x, then ddx(f(ln(x2)))=

  • 6((ln x)2+1)x

  • 3(ln x)2+3x

  • 3(ln x)2+3

  • 6(4(ln x)2+1)x

4
1 point

Let f be a differentiable function such that f(2)=−1, f(3)=2, f'(2)=−3 and f'(3)=−2.

The function g is differentiable and g(x)=f−1(x) for all x.

What is the value of g'(2)?

  • −13

  • −1

  • −12

  • 3

5
1 point

If f(x)=ln(2x−3+e−x), then f''(0)=

  • −13

  • −34

  • −12

  • −14

6
1 point

An equation of the line tangent to the graph of f(x)=x(2−3x)4 at the point (1, 1) is

  • y=−12x+13

  • y=−11x+12

  • y=11x−10

  • y=13x−12

7
1 point

If f(x)=(x+3)(x2−1)3, then f'(x)=

  • 6x(x2−1)2

  • 6x(x+3)(x2−1)2

  • (x2−1)2(x2+3x+8)

  • (x2−1)2(7x2+18x−1)