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AP Calculus AB: 6.2.2 Applying Implicit Differentiation

Mathematics9 CardsCreated 8 months ago

This flashcard set demonstrates how to apply implicit differentiation to find derivatives of relations and use them to determine the equations of tangent lines to curves. It emphasizes the use of derivative rules like the product and chain rule, and shows how to substitute points into the derivative to compute the slope for point-slope form equations of lines.

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Applying Implicit Differentiation

  • 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.

  • To find the equation of a line tangent to a curve, you need a point on the line and the slope of the line. To find the slope of the line, you may need to substitute both the x-value and the y-value of the point into the derivative.

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Key Terms

Term
Definition

Applying Implicit Differentiation

  • Find the derivative of a relation by differentiating each side of its equation implicitly and solving for the derivative as an unknown. Thi...

note

  • To find the derivative of this relation you must use
    implicit differentiation.

  • Take the derivative of both sides of the e...

What is the equation of the line tangent to the curve xy = 4 at the point (2, 2)?

( y − 2) = −1 (x − 2)

Suppose a curve is defined by the equation x^2/3+ y^2/3=4.Which set represents all the points on the curve where the line tangent to the curve has slope m=−1?

{(√8,√8),(−√8,−√8)}

Suppose a curve is defined by the equation y/y−x=x^2−1.Find dx/dy.

dx/dy=x/y−2x(y−x)^2


What is the equation of the line tangent to the curve

x^ 2 + y^ 2 = 25 at the point (0, −5)?

y = −5

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AP Calculus AB: 6.2.2 Applying Implicit Differentiation
TermDefinition

Applying Implicit Differentiation

  • 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.

  • To find the equation of a line tangent to a curve, you need a point on the line and the slope of the line. To find the slope of the line, you may need to substitute both the x-value and the y-value of the point into the derivative.

note

  • To find the derivative of this relation you must use
    implicit differentiation.

  • Take the derivative of both sides of the equation.

  • Notice that you must use the chain rule and the product rule to find the derivative.

  • Now substitute the point of tangency into the derivative to find the slope. Notice that you must substitute both the x-value and the y-value.

  • Use the point-slope formula to find the equation of the tangent line.

  • This implicit equation will require several different
    differentiation rules to differentiate.

  • It is a good idea to differentiate each term as a side-problem first and then to combine all of the results at the end of the problem.

What is the equation of the line tangent to the curve xy = 4 at the point (2, 2)?

( y − 2) = −1 (x − 2)

Suppose a curve is defined by the equation x^2/3+ y^2/3=4.Which set represents all the points on the curve where the line tangent to the curve has slope m=−1?

{(√8,√8),(−√8,−√8)}

Suppose a curve is defined by the equation y/y−x=x^2−1.Find dx/dy.

dx/dy=x/y−2x(y−x)^2


What is the equation of the line tangent to the curve

x^ 2 + y^ 2 = 25 at the point (0, −5)?

y = −5

Suppose a curve is defined by the equation x^2−xy+y^3=8.Find dx/dy.

dx/dy=x−3y^2/2x−y

What is the equation of the line tangent to the curve x^ 2 + y^ 2 = 100 at the point (6, 8)?

y − 8 = −3/4 (x − 6)

Suppose a curve is defined by the equation
(x+1)^2/25+(y−1)^2/16=1.Which set represents all the points on the curve where the line tangent to the curve has slope m=1?

{(−√41+25/√41,√41−16/√41) , (−√41−25/√41,√41+16/√41)}