设y=xsinx,求dx分之dy-搜答案-智慧作业帮

是y=linx²么dy/dx=1/x².（x²）'=1/x².(2x)=2x/x²dy/dx|x=1=2*1/1²=2

再答：再问：再问：拜托了

设y+xe^y=1,求dy/dx

隐函数求导问题把有y看成x函数两端求导y'+e^y+xe^y*y'=0解出y'=-(e^y)/(1+x*e^y)OK?

x/y=ln(y/x)x(-1/y^2)y'+1/y=x/y(-y/x^2+y'/x)(1/y+x/y^2)y'=1/y+1/x[(y+x)/y^2]y'=(x+y)/xyy'=y/x

dx/dt=2tdy/dt=-sin(t)dy/dx=-sin(t)/2t同理:d²y/dx²=-cos(t)/2

dy/dx=[1-1/(1+t²)]/[2t/(1+t²)]=t/2d²y/dx²=(1/2)*dt/dx=(1/2)/(dx/dt)=(1/2)/[2t/(1

设y=2arctan(y/x),求dy/dx,d²y/dx².设F(x,y)=y-2arctan(y/x)=0,则dy/dx=-(∂F/∂x)/(ͦ

y'=x'*sinx^2+x*(sinx^2)'=sinx^2+x*cosx^2*(x^2)'=sinx^2+x*cosx^2*2x=sinx^2+2x^2*cosx^2再问：您好，我算的步骤是：y'

这是隐函数求导,y=xe^y,两边分别对x求导dy/dx=e^y+xe^y(dy/dx)dy/dx=e^y/(1-xe^y)在对上式求导d^2(y)/dx^2=[(dy/dx)e^y(1-xe^y-e

dx/dt=-e^tdy/dt=1-e^-tdy/dx=(dy/dt)/(dx/dt)=[e^(-t)-1]/e^td(dy/dt)/dt=-e^(-t)*e^t-e^t*[e^(-t)-1]/e^2

y'=1/f(x)*f'(x)=f'(x)/f(x)y''=f''(x)f(x)-f'(x)^2/f(x)^2

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dy/dx=(dy/dt)/(dx/dt)=-sint/2td²y/dx²=d(dy/dx)/dx=[d(dy/dx)/dt]/(dx/dt)=d(-sint/2t)/dt/2t=

y=sin(xy)dy/dx=cos(xy)*y=ycos(xy)d²y/dx²=-ysin(xy)*y=-y²sin(xy)

设y=x3+xsinx求导数y

y'=(X^3)'+(xsinx)'=3x^2+(x)'sinx+x(sinx)'=3x^2+sinx+xcosx

偶函数.f(x)=xsinxf(-x)=(-x)sin(-x)=-x*(-sinx)=xsinx=f(x)符合偶函数定义

y=cotx-xsinxy'=-(cscx)^2-sinx-xcosx再问：�й��û��лл再答：d/dx(cotx)=-(cscx)^2d/dx(xsinx)=xd/dx(sinx)+sinxd/

dy+d(x*e^y)=d(1)dy+xd(e^y)+e^ydx=0dy+xe^ydy+e^ydx=0(xe^y+1)dy=-e^ydxdy/dx=-e^y/(xe^y+1)

dx/dt=[t*1/t-2t(1+lnt)/t^4=(-1-2lnt)/t³dy/dt=[t*2/t-(3+2lnt)]/t²=(t-3-2lnt)/t²dy/dx=(

x=1/t²+lntdx/dt=-2/t³+1/t=(t²-2)/t³t=3/t+2sintdy/dx=-3/t²+2cost=(2t²co