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Engineering Calculus Notes 360

# Engineering Calculus Notes 360 - y △ x so for some t ′...

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348 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION Now, y x f = g (1) g (0) and the Mean Value Theorem applied to g ( t ) tells us that for some ˜ t (0 , 1), this diFerence = g ( ˜ t ) = b ∂f ∂y ( x 0 + x,y 0 + ˜ t y ) ∂f ∂y ( x 0 ,y 0 + ˜ t y ) B y or, writing ˜ y = y 0 + ˜ t y , and noting that ˜ y lies between y 0 and y 0 + y , we can say that y x f = b ∂f ∂y ( x 0 + x, ˜ y ) ∂f ∂y ( x 0 , ˜ y ) B y where ˜ y is some value between y 0 and y 0 + y . But now apply the Mean Value Theorem to h ( t ) = ∂f ∂y ( x 0 + t x, ˜ y ) with derivative h ( t ) = 2 f ∂x∂y ( x 0 + t x, ˜
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Unformatted text preview: y ) △ x so for some t ′ ∈ (0 , 1) b ∂f ∂y ( x + △ x, ˜ y ) − ∂f ∂y ( x , ˜ y ) B = h (1) − h (0) = h ′ ( t ′ ) = ∂ 2 f ∂x∂y ( x + t ′ △ x, ˜ y ) △ x and we can say that △ y △ x f = b ∂f ∂y ( x + △ x, ˜ y ) − ∂f ∂y ( x , ˜ y ) B △ y = ∂ 2 f ∂x∂y (˜ x, ˜ y ) △ x △ y...
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