AP Calculus AB Practice Quiz: Derivatives of $\\cos x$, $\\sin x$, $e^x$, and $\\ln x$

Correct Answer: A

This question tests the knowledge of a fundamental derivative rule. The derivative of the sine function is the cosine function. Therefore, d/dx(sin(x)) = cos(x). [cite: 1784]

Correct Answer: B

This question tests the knowledge of a fundamental derivative rule. The derivative of the cosine function is the negative sine function. Therefore, d/dx(cos(x)) = -sin(x). [cite: 1784]

Correct Answer: C

The derivative of the natural exponential function e^x is unique in that it is equal to the function itself. d/dx(e^x) = e^x. The derivatives of the other functions are d/dx(ln(x)) = 1/x, d/dx(sin(x)) = cos(x), and d/dx(cos(x)) = -sin(x). [cite: 1785]

Correct Answer: B

This question tests the knowledge of a fundamental derivative rule. The derivative of the natural logarithmic function ln(x) is 1/x. Therefore, d/dx(ln(x)) = 1/x. [cite: 1785]

Correct Answer: C

The expression is the limit definition of the derivative, lim(h→0) [f(a+h) - f(a)] / h. By comparing the given limit to this definition, we can identify the function as f(x) = cos(x) and the point as a = π/3. [cite: 1790, 1791]

Correct Answer: A

This limit represents the derivative of the function f(x) = e^x at the point x=2. The derivative of f(x) = e^x is f'(x) = e^x. Therefore, the value of the limit is f'(2) = e^2. This is an application of recognizing the definition of a derivative to evaluate a limit. [cite: 1794]

Correct Answer: A

Using the sum/difference and constant multiple rules for derivatives, we differentiate term by term. The derivative of 2e^x is 2e^x, and the derivative of -5sin(x) is -5cos(x). Thus, f'(x) = 2e^x - 5cos(x). [cite: 1780, 1781]

Correct Answer: B

The expression is the definition of the derivative of f(x) = ln(x) at the point a=1. The derivative of ln(x) is f'(x) = 1/x. Evaluating this at a=1 gives f'(1) = 1/1 = 1. [cite: 1794]

Correct Answer: C

This limit can be rewritten by noting that sin(π) = 0. The expression is equivalent to lim(h→0) [sin(π + h) - sin(π)] / h. This is the definition of the derivative of f(x) = sin(x) at x=π. The derivative is f'(x) = cos(x). Therefore, the value of the limit is f'(π) = cos(π) = -1. [cite: 1794]

Correct Answer: C

The slope of the tangent line is given by the derivative of the function at that point. The derivative of y = cos(x) is dy/dx = -sin(x). At x = π/2, the slope is -sin(π/2) = -1. [cite: 1781]

Correct Answer: B

The definition of the derivative of a function f at a point x=a is lim(h→0) [f(a+h) - f(a)] / h. For the function f(x) = ln(x), this definition becomes lim(h→0) [ln(a+h) - ln(a)] / h. Option A represents the derivative function f'(x), not the derivative at a specific point a. Option C is missing the '-a' in the denominator for the alternate form of the derivative. Option D incorrectly places the h outside the function. [cite: 1790]

Correct Answer: C

This limit can be recognized as the derivative of a function at a point. We can rewrite it as lim(x→0) [cos(x) - cos(0)] / (x - 0), since cos(0)=1. This is the alternate form of the definition of the derivative of the function f(x) = cos(x) at the point x=0. The derivative is f'(x) = -sin(x). Evaluating at x=0 gives f'(0) = -sin(0) = 0. [cite: 1794]

Correct Answer: D

First, find the derivatives of both functions. f'(x) = d/dx(ln(x)) = 1/x and g'(x) = d/dx(e^x) = e^x. We need to find a value of x such that f'(x) = g'(x), which means 1/x = e^x. The domain of f(x) = ln(x) is x > 0. For x > 0, the function y=e^x is always positive and increasing, while the function y=1/x is positive and decreasing. A graphical analysis shows these two functions never intersect. Therefore, there is no such value of x. [cite: 1781]