Calc03_7 - 3.7 Implicit Differentiation By...

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3.7 Implicit Differentiation By leonardogillesfleur
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y 1 = ξ ψ 2 =- ξ y 1 = 2 3 9-ξ 2 ψ 2 =- 2 3 9-ξ 2 y 1 = 2 ψ 2 =- 2 y 1 = 9-ξ 2 ψ 2 =- 9-ξ 2 y 1 = 2ξ+ 3-ξ 2 ψ 2 =- 2ξ+ 3-ξ 2
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y = 4 x - y +2 xy x 2 y = y + x cos x sin x - x y = x y 2 x 2 - y + x
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x 3 2 5 6 x - 1 2 - 5 6
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2 2 1 x y + = This is not a function, but it would still be nice to be able to find the slope. 2 2 1 d d d x y dx dx dx + = Do the same thing to both sides. 2 2 0 dy x y dx + = Note use of chain rule. 2 2 dy y x dx = - 2 2 dy x dx y - = dy x dx y = -
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2 2 sin y x y = + 2 2 sin d d d y x y dx dx dx = + This can’t be solved for y . 2 2 cos dy dy x y dx dx = + 2 cos 2 dy dy y x dx dx - = ( 29 2 2 cos dy x y dx = - 2 2 cos dy x dx y = - This technique is called implicit differentiation. 1 Differentiate both sides w.r.t. x . 2 Solve for . dy dx
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We need the slope. Since we can’t solve for y , we use implicit differentiation to solve for . dy
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This note was uploaded on 02/16/2012 for the course CALCULUS 0064 taught by Professor Waldron during the Fall '10 term at Broward College.

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Calc03_7 - 3.7 Implicit Differentiation By...

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