求图中导数求解化简过程(求导数，化简过程详细点)

求导数，化简过程详细点

y=arctan[(x^2+1)/(x^2-1)]^(1/2)

y'=1/{1+[(x^2+1)/(x^2-1)]}*{[(x^2+1)/(x^2-1)]^(1/2)}'

=[2x^2/(x^2-1)]*(1/2)[(x^2+1)/(x^2-1)]^(1/2-1)*[(x^2+1)/(x^2-1)]'

=[x^2/(x^2-1)]*[(x^2+1)/(x^2-1)]^(-1/2)*[(x^2+1)'(x^2-1)-(x^2+1)(x^2-1)']/(x^2-1)^2

=[x^2/(x^2-1)]*[(x^2+1)/(x^2-1)]^(-1/2)*[2x(x^2-1)-(x^2+1)2x]/(x^2-1)^2

=[x^2/(x^2-1)]*[(x^2+1)/(x^2-1)]^(-1/2)*(-4x)/(x^2-1)^2

=-4x^3(x^2+1)^(-1/2)*(x^2-1)^(-7/2)

=-4x^3/[(x^2+1)^(1/2)*(x^2-1)^(7/2)]

y=ln(x+√x^2+a^2)

那么y对x 求导得到

y'=1/(x+√x^2+a^2) *(x+√x^2+a^2)'

而显然

(x+√x^2+a^2)'

=1 +2x/2√x^2+a^2

=(x+√x^2+a^2)/√x^2+a^2

所以就解得

y'=1/(x+√x^2+a^2) *(x+√x^2+a^2)/√x^2+a^2

=1/√x^2+a^2

隐函数求导：arctan(y/x)=ln√(x²+y²)，求dy/dx

解：设F(x，y)=arctan(y/x)-ln√(x²+y²)=arctan(y/x)-(1/2)ln(x²+y²)=0

则dy/dx=-(∂F/∂x)/(∂F/∂y)...........(1)；

其中∂F/∂x=-(y/x²)/[1+(y/x)²]-(1/2)(2x)/(x²+y²)=-y/(x²+y²)-x/(x²+y²)=-(x+y)/(x²+y²)

∂F/∂y=(1/x)/[1+(y/x)²]-(1/2)[2y/(x²+y²)]=x/(x²+y²)-y/(x²+y²)=(x-y)/(x²+y²)

代入(1)式即得dy/dx=[(x+y)/(x²+y²)]/[(x-y)/(x²+y²)]=(x+y)/(x-y)

【看不清楚你写的，请自己对照。】

把它看做-4乘以x的-1次方，然后用公式。