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

Correct Answer: A

This question assesses the ability to calculate the derivative of a familiar function. The derivative of $\sin x$ is a specific rule. According to the differentiation rules for trigonometric functions, $\frac{d}{dx}(\sin x) = \cos x$. [cite: 1780, 1784]

Correct Answer: B

This question assesses the ability to calculate the derivative of a familiar function involving a constant multiple. The derivative of $e^x$ is $e^x$. The constant multiple rule states that $\frac{d}{dx}(c \cdot f(x)) = c \cdot f'(x)$. Therefore, $\frac{d}{dx}(5e^x) = 5 \cdot \frac{d}{dx}(e^x) = 5e^x$. [cite: 1781, 1785]

Correct Answer: A

This question requires recognizing an expression for the definition of the derivative, $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$, to determine a limit. Here, the function is $f(x) = e^x$ and the point is $a=3$. The limit is therefore equal to $f'(3)$. Since $f'(x) = e^x$, the value of the limit is $f'(3) = e^3$. [cite: 1794]

Correct Answer: C

This question assesses the ability to calculate derivatives of familiar functions using the sum/difference rule. The derivative of $\ln x$ is $\frac{1}{x}$ and the derivative of $\cos x$ is $-\sin x$. Therefore, $\frac{d}{dx}(\ln x - \cos x) = \frac{d}{dx}(\ln x) - \frac{d}{dx}(\cos x) = \frac{1}{x} - (-\sin x) = \frac{1}{x} + \sin x$. [cite: 1780, 1784, 1785]

Correct Answer: D

This question requires the ability to interpret a limit as a definition of a derivative. The limit definition of the derivative is $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$. By comparing the given expression to this definition, we can identify that $f(x) = \cos x$ and $a = \frac{\pi}{2}$. [cite: 1790, 1791]

Correct Answer: B

This question requires recognizing an expression for the definition of the derivative to determine a limit. The expression can be rewritten as $\lim_{h \to 0} \frac{\ln(1+h) - \ln(1)}{h}$, since $\ln(1) = 0$. This is the definition of the derivative of the function $f(x) = \ln x$ at the point $x=1$. The derivative of $f(x) = \ln x$ is $f'(x) = \frac{1}{x}$. Therefore, the value of the limit is $f'(1) = \frac{1}{1} = 1$. [cite: 1794]

Correct Answer: B

The slope of the tangent line at a point is given by the value of the derivative at that point. The derivative of $y = \cos x$ is $\frac{dy}{dx} = -\sin x$. To find the slope at $x = \pi$, we evaluate the derivative at that point: $-\sin(\pi) = 0$. [cite: 1780, 1784]

Correct Answer: B

This question requires interpreting a limit as a definition of a derivative. The definition of the derivative of $f(x)$ at $x=a$ is $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$. For $f(x) = e^x$ at $a=0$, this becomes $\lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{e^{h} - e^0}{h} = \lim_{h \to 0} \frac{e^h - 1}{h}$. [cite: 1790, 1791]

Correct Answer: A

This question assesses the ability to calculate the derivative of a familiar function and evaluate it at a specific point. First, find the derivative of $f(x) = \ln x$, which is $f'(x) = \frac{1}{x}$. Then, substitute $x = e^2$ into the derivative: $f'(e^2) = \frac{1}{e^2} = e^{-2}$. [cite: 1781, 1785]

Correct Answer: A

This question requires recognizing an expression for the definition of the derivative to determine a limit. Note that $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$. The expression is the definition of the derivative of $f(x) = \sin x$ at $x = \frac{\pi}{3}$. The derivative is $f'(x) = \cos x$. Evaluating at $x = \frac{\pi}{3}$ gives $f'(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$. [cite: 1794]

Correct Answer: C

This question assesses the ability to calculate the derivative of a familiar function. The derivative of $\cos x$ is a specific rule. According to the differentiation rules for trigonometric functions, $\frac{d}{dx}(\cos x) = -\sin x$. [cite: 1780, 1784]

Correct Answer: C

The rate of change of a function at a point is given by its derivative at that point. The derivative of $f(x) = e^x$ is $f'(x) = e^x$. To find the rate of change at $x = \ln(4)$, we evaluate $f'(\ln(4)) = e^{\ln(4)}$. Since the exponential and natural logarithm functions are inverses, $e^{\ln(4)} = 4$. [cite: 1781, 1785]

Correct Answer: D

This question requires interpreting a limit as a definition of a derivative. One form of the definition is $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$. The given limit is $\lim_{x \to 0} \frac{\cos x - 1}{x}$. Since $\cos(0) = 1$, we can rewrite the limit as $\lim_{x \to 0} \frac{\cos x - \cos(0)}{x-0}$. Comparing this to the definition, we see that $f(x) = \cos x$ and $a=0$. Thus, the limit represents the derivative of $\cos x$ at $x=0$. [cite: 1790, 1791]