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Questions and Answers
What is the coefficient of $30 ext{sin}(3x) ext{cos}(3x)(4 + ext{sin}^2(3x))^4$ when finding the derivative of $y = (4 + ext{sin}^2(3x))^5$?
What is the coefficient of $30 ext{sin}(3x) ext{cos}(3x)(4 + ext{sin}^2(3x))^4$ when finding the derivative of $y = (4 + ext{sin}^2(3x))^5$?
- 25
- 15
- 5
- 30 (correct)
Which expression correctly represents the relationship between $x$ and $y$ in $x^2 + ext{cot}(xy) = 17$ when finding its derivative?
Which expression correctly represents the relationship between $x$ and $y$ in $x^2 + ext{cot}(xy) = 17$ when finding its derivative?
- $2x + ext{csc}^2(xy)(y + xy') = 0$
- $2x - ext{csc}^2(xy)(y + xy') = 0$ (correct)
- $2x + ext{sec}^2(xy)(y + xy') = 0$
- $2x - ext{sec}^2(xy)(y + xy') = 0$
What is the product of derivatives for the function $y = ext{cot}^{-1}(e^{5x})$?
What is the product of derivatives for the function $y = ext{cot}^{-1}(e^{5x})$?
- $ rac{-5e^{5x}}{1 + e^{10x}}$ (correct)
- $ rac{5e^{5x}}{1 + e^{10x}}$
- $ rac{-e^{5x}}{1 + e^{10x}}$
- $ rac{-5e^{5x}}{1 + e^{5x}}$
For the function $y = (4 + ext{sin}^2(3x))^5$, what is the final form of its derivative?
For the function $y = (4 + ext{sin}^2(3x))^5$, what is the final form of its derivative?
Flashcards
Derivative of a Square Root Function
Derivative of a Square Root Function
The derivative of a function involving a square root is found using the chain rule. The derivative of the outer function (square root) is multiplied by the derivative of the inner function (the expression under the square root).
Chain Rule for Derivatives
Chain Rule for Derivatives
The derivative of a composite function is found by multiplying the derivative of the outer function by the derivative of the inner function. The derivative of the outer function is applied to the inner function, and then multiplied by the derivative of the inner function. The derivative of the inner function is often easier to determine once the outer function has been differentiated.
Implicit Differentiation
Implicit Differentiation
Implicit differentiation is used to find the derivative of a function that is not explicitly defined in terms of x. The chain rule is applied to any term with y, and dy/dx is treated as a separate variable. The resulting equation can then be solved for dy/dx.
Derivative of Inverse Cotangent
Derivative of Inverse Cotangent
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Study Notes
AP Calculus Quiz 4 - Advanced Differentiation
- Problem Solving Techniques: Problems involve finding dy/dx using implicit differentiation and chain rule.
- Implicit Differentiation: Used for equations where y is not explicitly isolated. Requires applying the chain rule to terms involving y.
- Chain Rule: Essential for finding derivatives of composite functions, for instance, functions within square roots or trigonometric functions.
- Trigonometric Functions: Problems involve derivatives of sine, cosine, and potentially other trigonometric functions.
- Product Rule: Applicable when differentiating functions that are products or combinations of two functions.
- Quotient Rule: Applicable when differentiating functions that are quotients.
- Constant Rule: Applies when differentiating a constant value with respect to a variable.
- Power Rule: Applied when differentiating terms involving variables raised to powers.
- Example Problem 1: Finding dy/dx for y = √(4 + 2x³) requires using the chain rule and the power rule.
- Example Problem 2: Finding y' for y = (4 + sin² (3x))⁵. This requires combined use of chain rule, power rule, and trigonometric rules.
- Example Problem 3: Finding dy/dx for x² + cot(xy) = 17 using implicit differentiation. Requires applying the chain rule to the cotangent function.
- Example Problem 4: Finding the derivative y' for y = cot⁻¹(e^(sx)). Involves using the chain rule when differentiating an exponential function.
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Description
Test your knowledge on advanced differentiation techniques in AP Calculus. This quiz covers implicit differentiation, chain rule, and the application of various differentiation rules including product, quotient, and power rules. Ideal for students preparing for AP exams.
