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differentiation

發問:

1 Find dy/dx if 3x2y+xy2+1=0 2Find the derivative of growth curve y(t)=30e0.05t 6 Find y' if y=e(x^2+2x) 7 Find y' if y=(1n x)3 希望各位高手幫下我 最好詳細d,我驚我唔明 THX GRATITUDE 更新: 第一條係咁 Ans: -(6xy + y2)/(3x2 + 2xy) 但我想知步驟 希望可以詳細D THX 1. Find dy/dx if 3x2y + xy2 +1 = 0 Ans: -(6xy + y2)/(3x2 + 2xy)

最佳解答:

1.) Did you mean Differentiate 3x^2y + xy^2+1 = 0 in respect of x? If so, the result is 6xy + y^2 + 1 = 0 And do you need to Differentiate the result in respect of y? If so, the result is 6x + 2y = 0 2.) y(t) = 30e^(0.05t) Let u = 0.05t du/dt = 0.05 du = 0.05 dt dy(t) / du = 30e^(u) Sub in u and du dy(t) = 30 e^(0.05t) (0.05dt) dy(t)/dt = 1.5 e^(0.05t) 6.) y = e^(x^2+2x) Let u = x^2 + 2x du/dx = 2x+2 du = (2x + 2) dx y = e^u dy/du = e^u dy = e^u du Sub in u and du dy = e^(x^2+2x) (2x+2) dx dy/dx = (2x+2) e^(x^2+2x) 7.) y = (ln x)^3 Let u = ln x du/dx = 1/x du = 1/x dx y = u^3 dy/du = 3u^2 dy = 3u^2 du Sub in u and du dy = 3(ln x)^2 1/x dx dy/dx = (3/x) (ln x)^2 2008-10-28 20:42:17 補充: 應該用Produce rule即y=uv-->y`=uv`+vu` 3x^2 y+xy^2+1=0 Differentiate in respect of x d/dx (3x^2 y)+d/dx (xy^2)+d/dx (1) = 0 3 d/dx (x^2 y)+d/dx (xy^2)+0 = 0 3(x^2(dy/dx)+y(2x))+(xy(dy/dx)+(dx/dx)y^2) = 0 (3x^2)(dy/dx)+6xy+(xy)(dy/dx)+(1)(y^2) = 0 (3x^2)(dy/dx)+(xy)(dy/dx)=-6xy-y^2 2008-10-28 20:42:21 補充: (dy/dx)(3x^2+xy)=-(6xy+y^2) dy/dx=-(6xy + y^2)/(3x^2+xy)

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1. 3 x^2 y+x y^2+1=0 3(2xy+x^2 y')+(y^2+2xyy')=0 y'=-(6xy+y^2)/(3x^2+2xy) 2. y'(t)=30*0.05e^(0.05t)=1.5e^(0.05t) 6. y=e^(x^2+2x) y'=(2x+2)*e^(x^2+2x)=2(x+1)e^(x^2+2x) 7. y=(ln x)^3 y'=3(ln x)^2*1/x
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