Derivatives and Differentiation Quiz
Grade 12 · Math · Quiz
Test your mastery of differentiation rules including the Chain Rule, Product Rule, implicit differentiation, and related applications.
Questions
What is the derivative of f(x) = (x² + 1)⁴?
- 4(x² + 1)³
- 8x(x² + 1)³
- 4x(x² + 1)³
- 8(x² + 1)³
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Answer: 8x(x² + 1)³
Chain Rule: outer (u⁴)' = 4u³, inner (x²+1)' = 2x. So 4(x²+1)³ · 2x = 8x(x²+1)³.
Find f'(x) if f(x) = (2x + 3) / (x² - 1).
- (2(x² - 1) - 2x(2x + 3)) / (x² - 1)²
- (2x² - 2 - 4x² - 6x) / (x² - 1)²
- Both A and B are equivalent
- 2 / (2x)
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Answer: Both A and B are equivalent
Quotient Rule gives (2(x²-1) - (2x+3)(2x)) / (x²-1)². Expanding: (2x²-2-4x²-6x)/(x²-1)². Both options A and B simplify to the same expression.
Use implicit differentiation on x² + y² = 25. Find dy/dx.
- x/y
- -x/y
- y/x
- -y/x
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Answer: -x/y
Differentiating implicitly: 2x + 2y(dy/dx) = 0, so dy/dx = -2x/(2y) = -x/y.
If g(x) = e^(x² - 3x), then g'(x) is:
- e^(x² - 3x)
- (2x - 3)e^(x² - 3x)
- 2x · e^(x² - 3x)
- (x² - 3x)e^(x² - 3x - 1)
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Answer: (2x - 3)e^(x² - 3x)
Chain Rule on e^u where u = x²-3x: g'(x) = e^(x²-3x) · (2x-3).
At which x-value does the function f(x) = x³ - 3x² + 2 have a local maximum?
- x = 0
- x = 2
- x = -1
- x = 1
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Answer: x = 0
f'(x) = 3x²-6x = 3x(x-2). Critical points at x=0 and x=2. f''(x)=6x-6; f''(0)=-6 < 0 so x=0 is a local maximum.
The second derivative f''(x) of f(x) = 4x³ - 6x² + x is:
- 12x - 6
- 24x - 12
- 12x² - 12x + 1
- 8x - 6
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Answer: 24x - 12
f'(x) = 12x²-12x+1; f''(x) = 24x-12.
If f''(x) > 0 on an interval, then f is concave up on that interval.
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Answer: True
The second derivative test for concavity: f''(x) > 0 means the slope of f' is increasing, which corresponds to concave up (cup-shaped).
The derivative of f(x) = cos(x) is ___.
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Answer: -sin(x)
The derivative of cosine is negative sine: d/dx[cos x] = -sin x.
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