3_5 - Section 3.5: Implicit Differentiation Instructor:...

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Unformatted text preview: Section 3.5: Implicit Differentiation Instructor: Ms. Hoa Nguyen (nguyen@scs.fsu.edu) Implicit function The relation between x and y is not explicitly expressed as y = f (x). For examples, x2 + y 2 = 25 (a circle) or x3 + y 3 = 6xy (the folium of Descartes). The method of implicit differentiation This consists of differentiating both sides of the implicit function with respect to x and then solving the resulting equation for y . Example 1 of Section 3.5 (textbook): Example 2 of Section 3.5 (textbook): Example 3 of Section 3.5 (textbook): Example 4 of Section 3.5 (textbook): 1 Inverse Trigonometric Functions If a function f is one-to-one (i.e., passes the horizontal line test), then it has the inverse function f −1 . f −1 is also one-to-one and has its inverse f . For examples, the function f (x) = sin x, −2Π ≤ x ≤ Π has the inverse f −1 (x) = sin−1 x = 2 arcsin x. Remarks: 1) The graphs of f and f −1 are reflected about the line y = x. 2) Recall the definition of the arcsine function: y = sin−1 x ⇔ sin y = x and − π ≤ x ≤ π . 2 2 Inverse Trigonometric Functions: −Π 2 • f (x) = sin x, Π 2 ≤x≤ ⇔ f −1 (x) = sin−1 x = arcsin x. • f (x) = cos x, 0 ≤ x ≤ Π ⇔ f −1 (x) = cos−1 x = arccos x. • f (x) = tan x, −Π 2 ≤x≤ Π 2 ⇔ f −1 (x) = tan−1 x = arctan x. • f (x) = cot x, 0 ≤ x ≤ Π ⇔ f −1 (x) = cot−1 x. • f (x) = sec x, x ∈ [0, Π ) ∪ [Π, 3Π ) ⇔ f −1 (x) = sec−1 x, |x| ≥ 1. 2 2 • f (x) = csc x, x ∈ (0, Π ] ∪ (Π, 3Π ] ⇔ f −1 (x) = csc−1 x, |x| ≥ 1. 2 2 Derivatives of Inverse Trigonometric Functions • (sin−1 x) = √1 1 − x2 • (tan−1 x) = 1 1+x2 • (sec−1 x) = √1 x x2 − 1 (cos−1 x) = − √11 x2 − 1 (cot−1 x) = − 1+x2 1 (csc−1 x) = − x√x2 −1 Example 5 of Section 3.5 (textbook): 2 ...
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3_5 - Section 3.5: Implicit Differentiation Instructor:...

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