AP DC Calculus Final Exam Review - Derivatives

Find the slope of the tangent line to f(x) = -3x²-6x at x = 1.

• m = 0
• f'(x) = -6x - 6
• f'(x) = 6x
• m = -12

Find the derivative of f(x) = 6x³⁰ -2x¹⁵ + 4x³ - 2x + 1

• f'(x) = 18x²⁹ + 30x¹⁵ + 12x³ 
• f'(x) = 180x²⁹ - 30x¹⁴ + 12x² 
• f'(x) = 180x²⁹ - 30x¹⁴ + 12x² - 2
• f'(x) = 180x²⁹ - 30x¹⁴ + 12x² +1

If the position function for a particle is s(t) = -t² - t, what is the instantaneous velocity function for the particle? 

• v(t) = -2
• v(t) -2t - 1 
• v(t) = t³
• v(t) = -t

If the position of a particle is represented by s(t) = -t² + 1, what is its position at t = 1? 

• Position = 0
• Position = 1
• Position = 2
• Position = -1 

If the position of a particle is represented by s(t) = -t² + 1, what is its instantaneous velocity at t = 1?  

• Velocity = 0
• Velocity = 1
• Velocity = -1
• Velocity = -2

Find the derivative. f(x) = -8x^-3 + 5x - e^x

• f'(x) = -24x^-4 + 5 - e^x
• f'(x) = 24x^-4 + 5 - e^x
• f'(x) = 24x^-2 + 5 - e^x
• f'(x) = 24x^-4 + 5 - e^x-1

Find the derivative. f(x) = 1/(3x²) + 4x³

• f'(x) = 6x³ + 12x²
• f'(x) = 6x + 12x²
• f'(x) =  -2/(3x³) + 12x²
• f'(x) = -6/x² + 12x²

Find the derivative f(x) = (x² + 2x)⁵

• f'(x) = 5(2x+2)⁴
• f'(x) = 5(x² + 2x)⁴
•  f'(x) = 5(x² + 2x)⁴(2x)
• f'(x) = 5(x² + 2x)⁴(2x + 2)

Find the derivative f(x) = (3x + 5)⁴

• f'(x) = 12(3x)³
• f'(x) = 12(3x + 5)³
• f'(x) = (3)⁴
• f'(x) = (3x + 5)³

Find the derivative f(x) = secx

• f'(x) = tan²x
• f'(x) = sec²x
• f'(x) = tanx
• f'(x) = secxtanx