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Integral question
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最佳解答:
a. f'(x) = √[x^4 + 2] b.g'(x) = 5 + cos(x^2) g"(x) = - sin(x^2)*2x = -2x sin(x^2) c. h'(x) = 1/[1+x^24] - 1/2*1/[1+(x/2)^24] = 1/[1+x^24] - (2^23)/[x^24 + 2^24] y_1(x) = h(k(x)) y'_1(x) = h'(k(x))*k'(x) = {1/[1+sin^24 x] - (2^23)/[sin^24 x + 2^24]}*cos x d. y'_2(x) = f(x)g(x) y'_2(0) = f(0)g(0) = 0 e. y_3(x) = f(g(k(x))) y'_3(x) = f'(g(k(x)))*g'(k(x))*k'(x) k(x) = sin x, k(0)=0, k'(x) = cos x, k'(0) = 1 g'(x) = 5 + cos(x^2), g'(k(0)) = g'(0) = 5+cos0 = 6 g(k(0)) = g(0) = 0 f'(g(k(0))) = f'(0) = √[0^4 + 2] = √2 Therefore y'_3(0) = f'(g(k(0)))*g'(k(0))*k'(0) = √2*6*1 = 6√2 2009-01-29 13:37:43 補充： 最後一題計錯數，應該係計y'_3(π)： e. y_3(x) = f(g(k(x))) y'_3(x) = f'(g(k(x)))*g'(k(x))*k'(x) k(x) = sin x, k(π)=0, k'(x) = cos x, k'(π) = -1 g'(x) = 5 + cos(x^2), g'(k(π)) = g'(0) = 5+cos0 = 6 g(k(π)) = g(0) = 0 f'(g(k(π))) = f'(0) = √[0^4 + 2] = √2 Therefore y'_3(π) = f'(g(k(π)))*g'(k(π))*k'(π) = √2*6*-1 = -6√2
其他解答:
Integral question
發問:
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I need help with the following question: http://i611.photobucket.com/albums/tt192/kk-ke/5-co.jpg最佳解答:
a. f'(x) = √[x^4 + 2] b.g'(x) = 5 + cos(x^2) g"(x) = - sin(x^2)*2x = -2x sin(x^2) c. h'(x) = 1/[1+x^24] - 1/2*1/[1+(x/2)^24] = 1/[1+x^24] - (2^23)/[x^24 + 2^24] y_1(x) = h(k(x)) y'_1(x) = h'(k(x))*k'(x) = {1/[1+sin^24 x] - (2^23)/[sin^24 x + 2^24]}*cos x d. y'_2(x) = f(x)g(x) y'_2(0) = f(0)g(0) = 0 e. y_3(x) = f(g(k(x))) y'_3(x) = f'(g(k(x)))*g'(k(x))*k'(x) k(x) = sin x, k(0)=0, k'(x) = cos x, k'(0) = 1 g'(x) = 5 + cos(x^2), g'(k(0)) = g'(0) = 5+cos0 = 6 g(k(0)) = g(0) = 0 f'(g(k(0))) = f'(0) = √[0^4 + 2] = √2 Therefore y'_3(0) = f'(g(k(0)))*g'(k(0))*k'(0) = √2*6*1 = 6√2 2009-01-29 13:37:43 補充： 最後一題計錯數，應該係計y'_3(π)： e. y_3(x) = f(g(k(x))) y'_3(x) = f'(g(k(x)))*g'(k(x))*k'(x) k(x) = sin x, k(π)=0, k'(x) = cos x, k'(π) = -1 g'(x) = 5 + cos(x^2), g'(k(π)) = g'(0) = 5+cos0 = 6 g(k(π)) = g(0) = 0 f'(g(k(π))) = f'(0) = √[0^4 + 2] = √2 Therefore y'_3(π) = f'(g(k(π)))*g'(k(π))*k'(π) = √2*6*-1 = -6√2
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