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The derivative of a function ( f(x) ) at a point ( x ) is defined as the limit of the difference quotient: ( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} ). This topic is fundamental in calculus and appears in exams to test your understanding of limits and derivatives. Questions typically involve calculating derivatives using this definition and applying it to various functions.
This topic is tested in calculus exams, including AP Calculus, university-level calculus courses, and professional certifications like actuarial exams. It appears frequently and carries significant marks, often 10-20% of the total. It tests your ability to understand and apply the concept of limits to derive rates of change.
Intermediate
Question: Find the derivative of ( f(x) = x^2 ) using the limit definition.Step 1: Write the difference quotient: ( \frac{(x+h)^2 - x^2}{h} ).Step 2: Simplify the numerator: ( (x+h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2 ).Step 3: Divide by ( h ): ( \frac{2xh + h^2}{h} = 2x + h ).Step 4: Take the limit as ( h \to 0 ): ( \lim_{h \to 0} (2x + h) = 2x ).Answer: ( f'(x) = 2x ).
Question: Find the derivative of ( f(x) = \sqrt{x} ) using the limit definition.Step 1: Write the difference quotient: ( \frac{\sqrt{x+h} - \sqrt{x}}{h} ).Step 2: Rationalize the numerator: ( \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})} = \frac{h}{h(\sqrt{x+h} + \sqrt{x})} ).Step 3: Simplify: ( \frac{1}{\sqrt{x+h} + \sqrt{x}} ).Step 4: Take the limit as ( h \to 0 ): ( \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}} ).Answer: ( f'(x) = \frac{1}{2\sqrt{x}} ).
Question: Find the derivative of ( f(x) = \frac{1}{x} ) using the limit definition.Step 1: Write the difference quotient: ( \frac{\frac{1}{x+h} - \frac{1}{x}}{h} ).Step 2: Simplify the numerator: ( \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)} ).Step 3: Divide by ( h ): ( \frac{-h}{hx(x+h)} = \frac{-1}{x(x+h)} ).Step 4: Take the limit as ( h \to 0 ): ( \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x^2} ).Answer: ( f'(x) = -\frac{1}{x^2} ).
Question: What is the derivative of ( f(x) = 3x ) using the limit definition? Options: A) 2 B) 3 C) 4 D) 5 Correct Answer: B) 3 Explanation: ( f'(x) = \lim_{h \to 0} \frac{3(x+h) - 3x}{h} = \lim_{h \to 0} \frac{3h}{h} = 3 ).Why the Distractors Are Tempting: - A) Confuses the coefficient with the derivative.- C) Incorrect simplification.- D) Random guess.
Question: Find the derivative of ( f(x) = x^2 + 2x ) using the limit definition.Options: A) ( 2x ) B) ( 2x + 2 ) C) ( 2x + 1 ) D) ( 2x + 3 ) Correct Answer: B) ( 2x + 2 ) Explanation: ( f'(x) = \lim_{h \to 0} \frac{(x+h)^2 + 2(x+h) - (x^2 + 2x)}{h} = \lim_{h \to 0} \frac{2xh + h^2 + 2h}{h} = 2x + 2 ).Why the Distractors Are Tempting: - A) Forgets the linear term.- C) Incorrect simplification.- D) Random guess.
Question: What is the derivative of ( f(x) = \frac{1}{x^2} ) using the limit definition? Options: A) ( -\frac{2}{x^3} ) B) ( -\frac{1}{x^3} ) C) ( \frac{2}{x^3} ) D) ( \frac{1}{x^3} ) Correct Answer: A) ( -\frac{2}{x^3} ) Explanation: ( f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{hx^2(x+h)^2} = \lim_{h \to 0} \frac{-2xh - h^2}{hx^2(x+h)^2} = -\frac{2}{x^3} ).Why the Distractors Are Tempting: - B) Incorrect power.- C) Incorrect sign.- D) Incorrect power and sign.
Question: Find the derivative of ( f(x) = \sqrt{x+1} ) using the limit definition.Options: A) ( \frac{1}{2\sqrt{x+1}} ) B) ( \frac{1}{\sqrt{x+1}} ) C) ( \frac{1}{2(x+1)} ) D) ( \frac{1}{x+1} ) Correct Answer: A) ( \frac{1}{2\sqrt{x+1}} ) Explanation: ( f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+1} - \sqrt{x+1}}{h} = \lim_{h \to 0} \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}} = \frac{1}{2\sqrt{x+1}} ).Why the Distractors Are Tempting: - B) Forgets the factor of 2.- C) Incorrect form.- D) Incorrect form and factor.
Question: What is the derivative of ( f(x) = |x| ) using the limit definition? Options: A) ( \frac{x}{|x|} ) B) ( 1 ) C) ( -1 ) D) ( 0 ) Correct Answer: A) ( \frac{x}{|x|} ) Explanation: ( f'(x) = \lim_{h \to 0} \frac{|x+h| - |x|}{h} ). For ( x > 0 ), ( f'(x) = 1 ). For ( x < 0 ), ( f'(x) = -1 ). Thus, ( f'(x) = \frac{x}{|x|} ).Why the Distractors Are Tempting: - B) Only true for ( x > 0 ).- C) Only true for ( x < 0 ).- D) Incorrect, derivative is not 0.
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