7.2.1 Higher-Order Derivatives and Linear Approximation Flashcards Preview

AP Calculus AB > 7.2.1 Higher-Order Derivatives and Linear Approximation > Flashcards

Flashcards in 7.2.1 Higher-Order Derivatives and Linear Approximation Deck (14):
1

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.

2

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.

3

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

√3.9≈79/40

4

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

d^2y/dx^2=−16/y^3

5

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

y′′′=−27cos3x

6

Given f(x)=tanx, find f′′(x).

f′′(x)=2tanxsec^2x

7

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

ln(e+.1)≈10e+1/10e

8

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

√9.1≈181/60

9

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

d^2y/dx^2=−9/y^3

10

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

3√7.9≈239/120

11

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

f ′′′( x) = 72x

12

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

sin16/17π≈π/17

13

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

f ″(x) = 36x^ 2

14

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

d^3y/dx^3=−3/8y

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