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y=Cos1/x,求Dy/Dx

解:1/(1+y^2)y'=1/(1-(1/x)^2)^1/2*(-1/x^2) y'=(1+y^2)/x(x^2-1)^1/2。 答:y'=(1+y^2)/x(x^2-1)^1/2。

换元u=y-x,则dy/dx=du/dx+1,原方程化为du/dx=-cosu,分离变量du/cosu=-dx,两边积分ln(secu+tanu)=-lnx+lnC,所以x(secu+tanu)=C,代入u=y-x得通解x(sec(y-x)+tan(y-x))=C

你好!可以如图用复合函数求导公式得出答案。经济数学团队帮你解答,请及时采纳。谢谢!

利用隐函数定理. 记f(x,y)=cos(xy)-ln((x+1)/y)-1,则函数y=y(x)由f(x,y)=0确定,则dy/dx=-(df/dx)/(df/dy)=-(-sin(xy)*y-y/(x+1)/y)/(-sin(xy)*x-y/(x+1)*(x+1)/(-y^2))=(sin(xy)*y+1/(x+1))/(-sin(xy)*x+1/y),把x=0代入方程,解得y=1,代入dy...

求导,y'=ysinx+x*(y'sinx+y*cosx)-sin(x-1) 移项 y'-xy'sinx=ysinx+xycosx-sin(x-1) y'=(ysinx+xycosx-sin(x-1))/(1-xsinx)

不能作答

设u=√(x^2-1) dy/dx=(dy/du)(du/dx)=-cosu[x/√(x^2-1)] =-xcos[√(x^2-1)]/[√(x^2-1)]

记 O(0, 0), A(π,0), 补充线段 AO成封闭图形,则 I = ∫ = ∮ + ∫ 前者用格林公式, 顺时针加负号;后者 y = 0, dy = 0。 I = -∫∫ [(1+cosxy-xysinxy) - (cosxy-xysinxy)]dxdy + ∫xdx/(1+x^2) = -∫∫dxdy + (1/2)∫d(1+x^2)/(1+x^2) = - ∫sinxdx + (...

y=e^(1/x)+sin2x dy=e^(1/x)d(1/x)+cos2xd2x =e^(1/x)(-1/x^2)dx+2cos2xdx =[-e^(1/x)/x^2+2cos2x]dx

本题主要求y=x²的极坐标方程,即rsinθ=r²cos²θ,整理后为:r=sinθ/cos²θ则∫(0->1)dx∫(x^2->x)(x^2+y^2)^(-1/2)dy=∫[0->π/4]dθ∫[0->sinθ/cos²θ] (1/r)*rdr=∫[0->π/4]dθ∫[0->s...

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