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Back to FlashcardsMathematics / AP Calculus AB: 7.2.1 Higher-Order Derivatives and Linear Approximation

AP Calculus AB: 7.2.1 Higher-Order Derivatives and Linear Approximation

Mathematics14 CardsCreated 8 months ago

This flashcard set explores the concept of higher-order derivatives and how to use derivatives to create linear approximations for complex functions near known points. It includes practical examples of computing successive derivatives, using tangent lines for approximation, and applying these methods to estimate function values and solve implicit differentiation problems.

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Higher-Order Derivatives and Linear Approximation

  • You can find successive derivatives of a function by differentiating each result.

  • Derivatives can allow you to find a linear approximation for values of complicated functions near values that you know.

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Key Terms

Term
Definition

Higher-Order Derivatives and Linear Approximation

  • You can find successive derivatives of a function by differentiating each result.

  • note

    • A higher-order derivative is the derivative of a derivative. You can take as many higher-order derivatives as you like. In fact, some appli...

Use the linear approximation method from calculus to approximate the square root.
√3.9

√3.9≈79/40

Suppose x^2+y^2=16.Find d^2y/dx^2.

d^2y/dx^2=−16/y^3

Given the equation y = sin 3x, find y ′′′.

y′′′=−27cos3x

Given f(x)=tanx, find f′′(x).

f′′(x)=2tanxsec^2x

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AP Calculus AB: 7.2.1 Higher-Order Derivatives and Linear Approximation
TermDefinition

Higher-Order Derivatives and Linear Approximation

  • You can find successive derivatives of a function by differentiating each result.

  • Derivatives can allow you to find a linear approximation for values of complicated functions near values that you know.

note

  • A higher-order derivative is the derivative of a derivative. You can take as many higher-order derivatives as you like. In fact, some applications of calculus will require you to take an infinite number of higher-order derivatives.

  • The original derivative is called the first derivative. Each
    successive derivative is numbered one higher, so the next is the second derivative and the one following that is the third derivative.

  • Higher-order derivatives can also be described using Leibniz notation.

  • To indicate the second derivative of y with respect to x, put a superscripted 2 over the d in the numerator and over the y in the denominator.

  • Successive derivatives follow the same pattern.

  • The line tangent to a curve can be used to approximate values of the function. The tangent line is a good approximation close to the point of tangency because the tangent line behaves like the curve near that point and because lines are very easy to evaluate.

  • Using a line to estimate the value of a more complicated function is called a linear approximation.

  • To make a linear approximation, find the equation of the line tangent to the complicated curve at a value for the curve that you can evaluate, and close to the unknown value.

  • Once you have the equation of the line, plug the x-value of the unknown point into the equation of the line. The result is a good approximation of the original function.

Use the linear approximation method from calculus to approximate the square root.
√3.9

√3.9≈79/40

Suppose x^2+y^2=16.Find d^2y/dx^2.

d^2y/dx^2=−16/y^3

Given the equation y = sin 3x, find y ′′′.

y′′′=−27cos3x

Given f(x)=tanx, find f′′(x).

f′′(x)=2tanxsec^2x

Use the linear approximation method of calculus to approximate ln(e+.1).

ln(e+.1)≈10e+1/10e

Using the derivative to find a linear approximation, approximate √9.1.

√9.1≈181/60

x^2+y^2=9. Find d^2y/dx^2.

d^2y/dx^2=−9/y^3

Use the linear approximation method of calculus to approximate 3√7.9.

3√7.9≈239/120

Given the equation f (x) = 3x ^4, find f ′′′( x).

f ′′′( x) = 72x

Use the linear approximation method of calculus to approximate sin16π/17.

sin16/17π≈π/17

Given the equation f(x)=3x^4, find f′′(x).

f ″(x) = 36x^ 2

Suppose y^2=x^3.Find d^3y/dx^3.

d^3y/dx^3=−3/8y