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標題:

xe^y +ye^x = e

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Find dy/dx x e^y + y e^x = e e^(x+y) = e^x + e^y please show all the steps

最佳解答:

Question 1 x e^y + y e^x = e Differentiate w.r.t. x e^y + x e^y (dy/dx) + (dy/dx) e^x + y e^x = 0 (x e^y + e^x)(dy/dx) = -e^y - y e^x dy/dx = - (e^y + y e^x)/(x e^y + e^x) Question 2 e^(x+y) = e^x + e^y Differentiate w.r.t. x e^(x+y) (1+dy/dx) = e^x + e^y dy/dx e^(x+y) + e^(x+y) (dy/dx) = e^x + e^y (dy/dx) e^(x+y) - e^x = [e^y - e^(x+y)] (dy/dx) dy/dx = [e^(x+y) - e^x]/[e^y - e^(x+y)] dy/dx = [1 - e^(-y)]/[e^(-x) - 1]

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