www.nhft.net > 求 y=ln tAn x 的二阶导数,详细步骤,书上答案是%...

求 y=ln tAn x 的二阶导数,详细步骤,书上答案是%...

答案当然是对的, 但化简很麻烦,需要细心一些 ∂z/∂x =1/tan(x/y) * ∂[tan(x/y)]/∂x 而 ∂[tan(x/y)]/∂x =1/[cos(x/y)]^2 *∂[(x/y)]/∂x =1/y* 1/[cos(x/y)]^2 所以 ∂z/∂x =1/tan(x/...

x=ln(1+t²) y=t-arctant dx/dt=2t/(1+t²) dy/dt=1-1/(1+t²)=t²/(1+t²) ∴dy/dx=t/2 ∴d²y/dx²=½/[2t/(1+t²)]=(1+t²)/4t (y=t-arctanx,将x=ln(1+t²)代入,y=t-arctan[ln(1+t²)],计算...

解: dy/dx=(dy/dt)/(dx/dt) =[ln(1+t²)]'/(t-arctanx)' =[2t/(1+t²)]/[1- 1/(1+t²)] =2t/(1+t²-1) =2t/t² =2/t d²y/dx²=[d(2/t)/dt]/(dx/dt) =(-2/t²)/[1- 1/(1+t²)] =(-2/t²)/[(1+t²-...

(2 ArcTan[x])/(1 + x^2)

∂z/∂x=1/(1+y²/x²)*(-y/x²)=-y/(x²+y²) ∂z/∂y=1/(1+y²/x²)*1/x=x/(x²+y²) ∂²z/∂x²=y/(x²+y²)*2x=2xy/(x²+y²)² ∂&...

你的解法没问题啊,最后分子分母同时乘以cos^2(x+y)就可以化简了,过程参考:

因为y=ln|x|求导是1/x ,所以当做复合函数求导就可以了,自己算吧

对 √(x^2+y^2) = e^arctan(y/x) 两端取对数,得 (1/2)ln(x^2+y^2) = arctan(y/x), 两端求微分,得 (1/2)[(2xdx+2ydy)/(x^2+y^2)] = (xdy-ydx)/[1+(y/x)^2], 整理成 dy/dx = ……, 再求二阶导数……。

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