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AP Calculus AB: 6.2.1 Using Implicit Differentiation

Mathematics14 CardsCreated 3 months ago

This flashcard set focuses on applying implicit differentiation to find derivatives of equations involving multiple variables. It emphasizes using the chain rule and product rule within the process and highlights the importance of evaluating derivatives at specific points on the original relation by substituting both π‘₯ and 𝑦 values.

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Using Implicit Differentiation

β€’ Find the derivative of a relation by differentiating each side of its equation implicitly and solving for the derivative as an unknown. This process is called implicit differentiation.

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Key Terms

Term
Definition

Using Implicit Differentiation

β€’ Find the derivative of a relation by differentiating each side of its equation implicitly and solving for the derivative as an unknown. This proc...

note

  • Implicit differentiation often produces a derivative expressed in terms of more than one variable. When evaluating the slope of a line tang...

Given the equation 1βˆ’ln (xy) = e^y, find dy dx.

dy/dx=βˆ’y/xye^y+x

Given the equation xy=5,find dy/dx.

dy/dx=βˆ’y/x

Given the equation sinxy=1/2,find dy/dx.

dy/dx=βˆ’y/x

Given the equation x^2y+y^2x=0,find dy/dx.

dy/dx=βˆ’2xyβˆ’y^2/2xy+x^2

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AP Calculus AB: 6.2.1 Using Implicit Differentiation
TermDefinition

Using Implicit Differentiation

β€’ Find the derivative of a relation by differentiating each side of its equation implicitly and solving for the derivative as an unknown. This process is called implicit differentiation.

note

  • Implicit differentiation often produces a derivative expressed in terms of more than one variable. When evaluating the slope of a line tangent to a point of a relation, it is necessary to substitute both the x-value and the y-value of the point into the derivative.

  • Notice that you could substitute any values for x and y into the derivative. However, only ordered pairs of the original relation produce reasonable answers.

  • Here is a complicated-looking relation.

  • Find the derivative implicitly by taking the derivative of both sides of the implicit equation.

  • Now you can differentiate each term piece by piece.

  • Sometimes you will have to use different differentiation rules in the middle of a problem. Here the product and chain rules are both used.

  • Once you have differentiated each term, you can solve
    for dy/dx.

Given the equation 1βˆ’ln (xy) = e^y, find dy dx.

dy/dx=βˆ’y/xye^y+x

Given the equation xy=5,find dy/dx.

dy/dx=βˆ’y/x

Given the equation sinxy=1/2,find dy/dx.

dy/dx=βˆ’y/x

Given the equation x^2y+y^2x=0,find dy/dx.

dy/dx=βˆ’2xyβˆ’y^2/2xy+x^2

Given x/y=1/9,find dy/dx.

dy/dx=9

Suppose a curve is defined by the equation (6βˆ’x)y^2=x^3. What is the equation of the line tangent to the curve at (3, 3)?

y=2xβˆ’3

Given the equation2x+2y+xy^2=5,find dy/dx.

dy/dx=βˆ’2+y^βˆ’2/2βˆ’2xy^βˆ’3

Suppose a curve is defined by the equation y^2=x^3(2βˆ’x). What is the equation of the line tangent to the curve at (1,βˆ’1)?

y=βˆ’x

Given the equation x^2+3x=y^2+yβˆ’6,find dy/dx.

dy/dx=2x+3/2y+1

Given cos^3(sinxy)=k where k is some constant,find dxdy

dx/dy=βˆ’x/y

Given the equation xy=cotxy,find dy/dx.

dy/dx=βˆ’y/x

Given sin^3(cosxy)=k where k is some constant, find dy/dx.

dy/dx=βˆ’y/x