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I j h n _ k k b h g Z e v g u c i Z d _ l,. Let u = f (x) u = f (x) and v = g (x) v = g (x) be functions with continuous derivatives Then, the integrationbyparts formula for the integral involving these two functions is. At the point (u, v) = (0,3), where g(x, y) = x2 y2, 1 = e4u cos(u), Use the Chain Rule to evaluate the partial derivative y=e4u sin(v) (Use symbolic notation and fractions where needed) du (u,0) Get more help from Chegg Solve it with our calculus problem solver and calculator.

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A) u'(x) = g(x)f'(x) f(x)g'(x) this is the multiplication rule u'(1) = g(1)f'(1) f(1)g'(1) find these values by looking on the graph, remembering that the derivatives are just the slopes of the lines. Because u(x) = f(x)g(x) then (by the product rule, u'(x) = f '(x)g(x) f(x)g'(x) Looking at the graph we see that f(1) = 2 f '(1) = 2 g(1) = 1 g'(1) = 1. Ĕ̌ A } X ^ Z C g A v C x g x C g Ȃ ^ C } V E } P e B O y ̑I ѕ A ɁA g ̃f ^ ̑ ɁA Ĕ̌ A } X ^ Z C g A v C x g x C g Ȃ ^ C } V E } P e B O y Ɨގ ̎g ₷ ̈ 񏤍ނ ɂ́A т 񏤍ނ } j A ド L O ɉ āA ̏ W T C g A Z I ƁE d 烊 N Ă T r X Ă B.

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Because u(x) = f(x)g(x) then (by the product rule, u'(x) = f '(x)g(x) f(x)g'(x) Looking at the graph we see that f(1) = 2 f '(1) = 2 g(1) = 1 g'(1) = 1. WW2 B C x g u E E G X g t g v 14 N3 21 22 ̓ ԁA a ̎R { s ōs WW2 B e } ɂ q X g J Q C x g HOME. Вc @ l @ đ ό Y R ` đ s ۂ̓ 1360 Home > u 񂿂 v C X g M.

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