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Unit #2 Lesson #11 Higher Derivatives

The first time that a function is differentiated, the result is called the first derivative. If this derivative is differentiated again, the result is called the second derivative. This process can be repeated as long as it continues to be possible to differentiate. These derivatives (beyond the first derivative) are known as higher derivatives

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Notations for Higher Derivatives

Unit #2 Lesson #11 Higher Derivatives Notations for Higher Derivatives Second Derivative: f ” (x) or y ” or 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 Third derivative: f “‘ (x) or y “‘ or 𝒅 𝟑 𝒚 𝒅 𝒙 𝟑 These patterns continue as long as it is possible.

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Meanings of Derivatives

Unit #2 Lesson #11 Higher Derivatives Meanings of Derivatives The first derivative represents the instantaneous rate of change of a function. That is the slope of the tangent to the function at a point. The second derivative represents the rate at which the slope of the tangents to the function are changing as we move through the domain of the function

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Unit #2 Lesson #11 Higher Derivatives

Polynomial Functions EXAMPLE 1: If y = x3 – 2×2 + 5x – 7 then 𝑦 ′ = 𝑑𝑦 𝑑𝑥 =3 𝑥 2 −4𝑥+5 𝑦 ′′ = 𝑑 2 𝑦 𝑑 𝑥 2 =6𝑥−4 𝑦 ′′′ = 𝑑 3 𝑦 𝑑 𝑥 3 =6 Notice the change in degree from cubic to quadratic to linear. This pattern of reduction in degree is always true with polynomial functions, but the pattern becomes more complex with rational functions.

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Unit #2 Lesson #11 Higher Derivatives

EXAMPLE 2: Find the second derivative of 𝑓 𝑥 =−5𝑥 Unit #2 Lesson #11 Higher Derivatives First Derivative Second Derivative 𝑓 ′′ 𝑥 =0 𝑓 ′ 𝑥 =−5 EXAMPLE 3: Find the third derivative of f(x) = – 2×3 – 4×2 + 11x – 3 First Derivative Second Derivative f ′ (x) = – 6×2 – 8x + 11 f “(x) = –12x – 8 Third Derivative f “‘(x) = –12

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Unit #2 Lesson #11 Higher Derivatives

EXAMPLE 4: For the function f(x) = 12×4 – 3×2, find f ” (–2). f ′ (x) = 48×3 – 3x f “(x) = 144×2 – 6 f ” (–2) = 144(–2)2 – 6 = 570

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Unit #2 Lesson #11 Higher Derivatives

EXAMPLE 5: For f(x) = 4×3 – x + 3 find f ” (–1) 𝒇 ′ 𝒙 =𝟏𝟐 𝒙 𝟐 −𝟏 𝒇 ′′ 𝒙 =𝟐𝟒𝒙 𝒇 ′′ 𝒙 =𝟐𝟒 −𝟏 =−𝟐𝟒

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Unit #2 Lesson #11 Higher Derivatives

Chain Rule on Polynomial Function EXAMPLE 6: Find the second derivative of y = (x 3 + 2)5 Second Derivative With Product Rule First Derivative 𝒅𝒚 𝒅𝒙 =𝟓 𝒙 𝟑 +𝟐 𝟒 𝟑 𝒙 𝟐 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝒙 𝟑 +𝟐 𝟒 𝟑𝟎𝒙 +𝟏𝟓 𝒙 𝟐 (𝟒) 𝒙 𝟑 +𝟐 𝟑 𝟑 𝒙 𝟐 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝟑𝟎𝒙 𝒙 𝟑 +𝟐 𝟒 + 𝟏𝟖𝟎 𝒙 𝟒 𝒙 𝟑 +𝟐 𝟑 𝒅𝒚 𝒅𝒙 =𝟏𝟓 𝒙 𝟐 𝒙 𝟑 +𝟐 𝟒 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝟑𝟎𝒙 𝒙 𝟑 +𝟐 𝟑 𝒙 𝟑 +𝟐 𝟏 +𝟔 𝒙 𝟑 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝟑𝟎𝒙 𝒙 𝟑 +𝟐 𝟑 (𝟕 𝒙 𝟑 +𝟐)

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Unit #2 Lesson #11 Higher Derivatives

EXAMPLE 7: Find the second derivative of y = (2x 2 – 13)4 Unit #2 Lesson #11 Higher Derivatives First Derivative Second Derivative Use Product Rule 𝒅𝒚 𝒅𝒙 =𝟒 𝟐 𝒙 𝟐 −𝟏𝟑 𝟑 (𝟒𝒙) 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝟐 𝒙 𝟐 −𝟏𝟑 𝟑 𝟏𝟔 +(𝟏𝟔𝒙)(𝟑) 𝟐 𝒙 𝟐 −𝟏𝟑 𝟐 (𝟒𝒙) 𝒅𝒚 𝒅𝒙 =𝟏𝟔𝒙 𝟐 𝒙 𝟐 −𝟏𝟑 𝟑 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 =𝟏𝟔 𝟐 𝒙 𝟐 −𝟏𝟑 𝟑 +𝟏𝟗𝟐 𝒙 𝟐 𝟐 𝒙 𝟐 −𝟏𝟑 𝟐 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 =𝟏𝟔 𝟐 𝒙 𝟐 −𝟏𝟑 𝟐 𝟐 𝒙 𝟐 −𝟏𝟑 𝟏 +𝟏𝟐 𝒙 𝟐 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 =𝟏𝟔 𝟐 𝒙 𝟐 −𝟏𝟑 𝟐 𝟏𝟒 𝒙 𝟐 −𝟏𝟑

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Unit #2 Lesson #11 Higher Derivatives

EXAMPLE 8: Find the second derivative of 𝒚= 𝟐𝒙 𝟑𝒙+𝟏 First Derivative Second Derivative 𝒅𝒚 𝒅𝒙 = 𝟑𝒙+𝟏 𝟐 −𝟐𝒙(𝟑) 𝟑𝒙+𝟏 𝟐 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝟑𝒙+𝟏 𝟐 𝟎 −𝟐(𝟐)(𝟑𝒙+𝟏)(𝟑) 𝟑𝒙+𝟏 𝟒 𝒅𝒚 𝒅𝒙 = 𝟔𝒙+𝟐−𝟔𝒙 𝟑𝒙+𝟏 𝟐 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = −𝟏𝟐(𝟑𝒙+𝟏) 𝟑𝒙+𝟏 𝟒 = −𝟏𝟐 𝟑𝒙+𝟏 𝟑 𝒅𝒚 𝒅𝒙 = 𝟐 𝟑𝒙+𝟏 𝟐 OR OR 𝒅𝒚 𝒅𝒙 =𝟐 𝟑𝒙+𝟏 −𝟐 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 =−𝟒 𝟑𝒙+𝟏 −𝟑 𝟑 = −𝟏𝟐 𝟑𝒙+𝟏 𝟑

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Unit #2 Lesson #11 Higher Derivatives

Implicit Differentiation and Higher Order Derivatives EXAMPLE 9: Find 𝑑 2 𝑦 𝑑 𝑥 if 2 𝑥 3 −3 𝑦 2 =7 𝟔 𝒙 𝟐 −𝟔𝒚 𝒅𝒚 𝒅𝒙 =𝟎 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝒚 𝟐𝒙 − 𝒙 𝟐 𝒅𝒚 𝒅𝒙 𝒚 𝟐 −𝟔𝒚 𝒅𝒚 𝒅𝒙 =−𝟔 𝒙 𝟐 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝟐𝒙𝒚 𝒚 𝟐 − 𝒙 𝟐 𝒚 𝟐 × 𝒅𝒚 𝒅𝒙 𝒅𝒚 𝒅𝒙 = −𝟔 𝒙 𝟐 −𝟔𝒚 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝟐𝒙 𝒚 − 𝒙 𝟐 𝒚 𝟐 × 𝒙 𝟐 𝒚 𝒅𝒚 𝒅𝒙 = 𝒙 𝟐 𝒚 𝒅 𝟐 𝒚 𝒅 𝒙 𝟐 = 𝟐𝒙 𝒚 − 𝒙 𝟒 𝒚 𝟑

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