\[
\textbf{Question 1:}
\]
\[
\text{Which rule is used to differentiate the function } f(x) = g(x)h(x)?
\]
\[
\text{(a) } \text{Product Rule}, \quad \text{(b) } \text{Quotient Rule}, \quad \text{(c) } \text{Chain Rule}, \quad \text{(d) } \text{Power Rule}
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{The Product Rule states that if } f(x) = g(x)h(x),
\]
\[
\text{then } f'(x) = g'(x)h(x) + g(x)h'(x).
\]
\[
\text{Thus, the correct answer is } \boxed{\text{Product Rule}}.
\]
\[
\textbf{Question 2:}
\]
\[
\text{Find the derivative of } f(x) = x^n \text{ where } n \text{ is a constant.}
\]
\[
\text{(a) } nx^{n-1}, \quad \text{(b) } n^x, \quad \text{(c) } x^n, \quad \text{(d) } n
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{Using the Power Rule:} \quad \frac{d}{dx} x^n = nx^{n-1}.
\]
\[
\text{Thus, the correct answer is } \boxed{nx^{n-1}}.
\]
\[
\textbf{Question 3:}
\]
\[
\text{Which of the following is the derivative of } f(x) = \sin(x)?
\]
\[
\text{(a) } -\cos(x), \quad \text{(b) } \sin(x), \quad \text{(c) } \cos(x), \quad \text{(d) } -\sin(x)
\]
\[
\text{Answer: c}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{The derivative of } \sin(x) \text{ is } \cos(x).
\]
\[
\frac{d}{dx} \sin(x) = \cos(x).
\]
\[
\text{Thus, the correct answer is } \boxed{\cos(x)}.
\]
\[
\textbf{Question 4:}
\]
\[
\text{The derivative of } f(x) = \ln(x) \text{ is:}
\]
\[
\text{(a) } \frac{1}{x}, \quad \text{(b) } \ln(x), \quad \text{(c) } x, \quad \text{(d) } x^2
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{The derivative of } \ln(x) \text{ is } \frac{1}{x}.
\]
\[
\frac{d}{dx} \ln(x) = \frac{1}{x}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{x}}.
\]
\[
\textbf{Question 5:}
\]
\[
\text{Find the derivative of } f(x) = e^x.
\]
\[
\text{(a) } 1, \quad \text{(b) } x, \quad \text{(c) } e^x, \quad \text{(d) } x e^x
\]
\[
\text{Answer: c}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{The derivative of } e^x \text{ is } e^x.
\]
\[
\frac{d}{dx} e^x = e^x.
\]
\[
\text{Thus, the correct answer is } \boxed{e^x}.
\]
\[
\textbf{Question 6:}
\]
\[
\text{What is the derivative of } f(x) = \tan(x)?
\]
\[
\text{(a) } \sec^2(x), \quad \text{(b) } \sin(x), \quad \text{(c) } \cos(x), \quad \text{(d) } \sec(x)
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{The derivative of } \tan(x) \text{ is } \sec^2(x).
\]
\[
\frac{d}{dx} \tan(x) = \sec^2(x).
\]
\[
\text{Thus, the correct answer is } \boxed{\sec^2(x)}.
\]
\[
\textbf{Question 7:}
\]
\[
\text{Find the derivative of } f(x) = \frac{1}{x}.
\]
\[
\text{(a) } -x, \quad \text{(b) } x, \quad \text{(c) } -\frac{1}{x^2}, \quad \text{(d) } \frac{1}{x^2}
\]
\[
\text{Answer: c}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{Rewrite the function as } f(x) = x^{-1}.
\]
\[
\text{Differentiate using the Power Rule:} \quad \frac{d}{dx} x^n = n x^{n-1}.
\]
\[
\frac{d}{dx} x^{-1} = -1 \cdot x^{-2} = -\frac{1}{x^2}.
\]
\[
\text{Thus, the correct answer is } \boxed{-\frac{1}{x^2}}.
\]
\[
\textbf{Question 8:}
\]
\[
\text{Which rule is used to differentiate the composite function } f(x) = g(h(x))?
\]
\[
\text{(a) } \text{Product Rule}, \quad \text{(b) } \text{Chain Rule}, \quad \text{(c) } \text{Quotient Rule}, \quad \text{(d) } \text{Power Rule}
\]
\[
\text{Answer: B}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{The Chain Rule states that if } f(x) = g(h(x)),
\]
\[
\text{then } f'(x) = g'(h(x)) \cdot h'(x).
\]
\[
\text{Thus, the correct answer is } \boxed{\text{Chain Rule}}.
\]
\[
\textbf{Question 9:}
\]
\[
\text{The derivative of } f(x) = \cos(x) \text{ is:}
\]
\[
\text{(a) } -\sin(x), \quad \text{(b) } \sin(x), \quad \text{(c) } -\cos(x), \quad \text{(d) } \cos(x)
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{The derivative of } \cos(x) \text{ is } -\sin(x).
\]
\[
\frac{d}{dx} \cos(x) = -\sin(x).
\]
\[
\text{Thus, the correct answer is } \boxed{-\sin(x)}.
\]
\[
\textbf{Question 10:}
\]
\[
\text{If } f(x) = x^2 + 3x + 5, \text{ the derivative is:}
\]
\[
\text{(a) } 2x + 3, \quad \text{(b) } 2x + 5, \quad \text{(c) } 2x, \quad \text{(d) } 3x + 5
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\frac{d}{dx} (x^2 + 3x + 5) = 2x + 3.
\]
\[
\text{Thus, the correct answer is } \boxed{2x + 3}.
\]
\[
\textbf{Question 11:}
\]
\[
\text{Which of the following is the derivative of } f(x) = \sec(x)?
\]
\[
\text{(a) } \sec(x)\tan(x), \quad \text{(b) } \sec(x), \quad \text{(c) } \tan(x), \quad \text{(d) } -\sec(x)
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\frac{d}{dx} \sec(x) = \sec(x) \tan(x).
\]
\[
\text{Thus, the correct answer is } \boxed{\sec(x)\tan(x)}.
\]
\[
\textbf{Question 12:}
\]
\[
\text{The derivative of } f(x) = \arcsin(x) \text{ is:}
\]
\[
\text{(a) } \frac{1}{\sqrt{1 – x^2}}, \quad \text{(b) } \frac{1}{x^2}, \quad \text{(c) } \frac{1}{x\sqrt{1 – x^2}}, \quad \text{(d) } \sqrt{1 – x^2}
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 – x^2}}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{\sqrt{1 – x^2}}}.
\]
\[
\textbf{Question 13:}
\]
\[
\text{Find the derivative of } f(x) = \log_a(x).
\]
\[
\text{(a) } x \ln(a), \quad \text{(b) } \frac{1}{x}, \quad \text{(c) } \ln(a), \quad \text{(d) } \frac{1}{x \ln(a)}
\]
\[
\text{Answer: d}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\frac{d}{dx} \log_a(x) = \frac{1}{x \ln(a)}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{x \ln(a)}}.
\]
\[
\textbf{Question 14:}
\]
\[
\text{What is the derivative of } f(x) = \sqrt{x}?
\]
\[
\text{(a) } \frac{1}{2\sqrt{x}}, \quad \text{(b) } \frac{1}{\sqrt{x}}, \quad \text{(c) } \frac{1}{2x}, \quad \text{(d) } x^{\frac{1}{2}}
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{Rewrite as } f(x) = x^{\frac{1}{2}}.
\]
\[
\frac{d}{dx} x^{\frac{1}{2}} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{2\sqrt{x}}}.
\]
\[
\textbf{Question 15:}
\]
\[
\text{If } f(x) = \frac{2x^2 + 3x}{x^2 + 1}, \text{ what is its derivative?}
\]
\[
\text{(a) } \frac{(4x + 3)(x^2 + 1) – (2x^2 + 3x)(2x)}{(x^2 + 1)^2}, \quad \text{(b) } \frac{4x + 3}{x^2 + 1}, \quad \text{(c) } \frac{2x^2 + 3x}{x}, \quad \text{(d) } \frac{4x + 3}{x}
\]
\[
\text{Answer: A}
\]
Step by Step Solution
\[
\textbf{Solution:}
\]
\[
\text{Using the Quotient Rule: } \left( \frac{g(x)}{h(x)} \right)’ = \frac{g'(x)h(x) – g(x)h'(x)}{h(x)^2}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{(4x + 3)(x^2 + 1) – (2x^2 + 3x)(2x)}{(x^2 + 1)^2}}.
\]
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