15.3%2012%20-%2052

15.3%2012%20-%2052 - S E C T I O N 15.3 Partial Derivatives...

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SECTION 15.3 Partial Derivatives (ET Section 14.3) 641 12. z = x 2 + y 2 SOLUTION We compute z x ( x , y ) by treating y as a constant, and we compute z y ( x , y ) by treating x as a constant: x ( x 2 + y 2 ) = 2 x ; y ( x 2 + y 2 ) = 2 y 13. z = x 4 y 3 Treating y as a constant (to fnd z x )and x as a constant (to fnd z y ) and using Rules For DiFFerentiation, we get, x ( x 4 y 3 ) = y 3 x ( x 4 ) = y 3 · 4 x 3 = 4 x 3 y 3 y ( x 4 y 3 ) = x 4 y ( y 3 ) = x 4 · 3 y 2 = 3 x 4 y 2 14. z = x 4 y + xy 2 We obtain the Following partial derivatives: x ( x 4 y + 2 ) = 4 x 3 y + y 2 y ( x 4 y + 2 ) = x 4 + x · ( 2 y 3 ) = x 4 2 3 15. V = π r 2 h We fnd V r and V h : V r = r ( r 2 h ) = h r ( r 2 ) = h · 2 r = 2 hr V h = h ( r 2 h ) = r 2 16. z = x y Treating y as a constant we have x ± x y = 1 y x ( x ) = 1 y · 1 = 1 y We now fnd the derivative z y ( x , y ) , treating x as a constant: y ± x y = x · y ± 1 y = x · 1 y 2 = x y 2 . 17. z = x x y We diFFerentiate with respect to x , using the Quotient Rule. We get x ± x x y = ( x y ) x ( x ) x x ( x y ) ( x y ) 2 = ( x y ) · 1 x · 1 ( x y ) 2 = y ( x y ) 2 We now diFFerentiate with respect to y , using the Chain Rule: y ± x x y = x y ± 1 x y = x · 1 ( x y ) 2 y ( x y ) = x · 1 ( x y ) 2 · ( 1 ) = x ( x y ) 2 18. z = p 9 x 2 y 2 DiFFerentiating with respect to x , treating y as a constant, and using the Chain Rule, we obtain x ± q 9 x 2 y 2 = 1 2 p 9 x 2 y 2 x ( 9 x 2 y 2 ) = 2 x 2 p 9 x 2 y 2 = x p 9 x 2 y 2 We now diFFerentiate with respect to y , treating x as a constant: y ± q 9 x 2 y 2 = 1 2 p 9 x 2 y 2 y ( 9 x 2 y 2 ) = 2 y 2 p 9 x 2 y 2 = y p 9 x 2 y 2
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642 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 19. z = x p x 2 + y 2 SOLUTION We compute z x using the Quotient Rule and the Chain Rule: z x = 1 · p x 2 + y 2 x x p x 2 + y 2 ³ p x 2 + y 2 ´ 2 = p x 2 + y 2 x · 2 x 2 x 2 + y 2 x 2 + y 2 = x 2 + y 2 x 2 ( x 2 + y 2 ) 3 / 2 = y 2 ( x 2 + y 2 ) 3 / 2 We compute z y using the Chain Rule: z y = x y ( x 2 + y 2 ) 1 / 2 = x · ± 1 2 ( x 2 + y 2 ) 3 / 2 · 2 y = xy ( x 2 + y 2 ) 3 / 2 20. z = ( sin x )( sin y ) We obtain the following partial derivatives: x ( sin x sin y ) = sin y x sin x = sin y cos x y ( sin x sin y ) = sin x y sin y = sin x cos y 21. z = sin ( u 2 v) By the Chain Rule, d du sin ω = cos d and d d v sin = cos d d v . Applying this with = u 2 v gives u sin ( u 2 = cos ( u 2 u ( u 2 = cos ( u 2 · 2 u v = 2 u v cos ( u 2 ∂v sin ( u 2 = cos ( u 2 ( u 2 = cos ( u 2 · u 2 = u 2 cos ( u 2 22. z = tan x y By the Chain Rule, d dx tan u = 1 cos 2 u and d dy tan u = 1 cos 2 u .
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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15.3%2012%20-%2052 - S E C T I O N 15.3 Partial Derivatives...

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