dy tany

所以 $ cdot dy = dx$最终得到：$dy = frac{1}{tan^2 y} cdot dx$，也可以写作 $dy = cot^2 y cdot dx$综上所述，隐函数 $y$ 关于 $x$ 的微分 $dy...

设x=tany是直接函数,y属于(-pi/2,pi/2)则y=arctanx是它的反函数.函数x=tany在(-pi/2,pi/2)内单调可导(tany)'=sec^2y有反函数求导公式dy/dx=1/(dx/dy)得(arctanx...

如何用定义证明arctanx的导数？利用如下基本事实：arctanxarctany=arctanxy1+xy;arctanxx(x→0).于是有(arc...

x \right)^\prime &= \frac{1}{{{\left( \sin y \right)^\prime }}} &&\color{Red}{(\frac{dx}{dy}=\frac{1}{\frac{dy...

sinycosy=sin(2y)/2=tany/[1+(tany)^2]=u/(1+u^2).于是，x^2d^2y/dx^2+2x^2(tany)(dy/dx)^2+xdy/dx-sinycosy =(d^2u/dt^2-u)/(1+u^2).从...

tany=x siny/cosy=x siny=xcosy (siny)^2+(cosy)^2=1 即：(1+x^2)(cosy)^2=1 (cosy)^2=1/(1+x^2)所以y'=1/(1+x^2)dy=y'dx=[1/(1+x^2)...

u=tany,x=e^t.du=(secy)^2dy=[1+(tany)^2]dy=(1+u^2)dy,dy=du/(1+u^2), dx=e^tdt.dy/dx=1/[e^t(1+u^2)]du/dt,d^2y/dx^2=d(dy/dx)/...

把y的导数写成dy/dx，然后再把tany除到等号的左边那么就完成分离了：dy/tany=tanxdx

令x=a*tany => dx=a*sec²y dy tany=x/a，siny=x/√(x²+a²)，cosy=a/√(x²+a²)∫dx/(x²+a²)²= ...