y=loge(x^2+y^2),那么dy/dx在(1,0)的值是?

y=loge(x^2+y^2),那么dy/dx在(1,0)的值是? 1.Find a sequence of transformations which transforms the graph of y=loge (x) （以e为底x的对数）to the graph of y=-2loge (2-x) (打不出来和上边一样 见谅).2.Sketch the graph of y=-2loge (2-x)3.Solve the equation -2loge (2-x)=10 for x xy平面上的曲线y=loge(1-x^2)求[-0.5,0.5]上的曲线长. 请用 隐函数微分帮忙算出dy/dx x^3* y^3*(1-loge|x|)=1 *为乘号 y=x^2x,求dy (x^2)dy+（y^2)dx=dx-dy y=e^2x,dy= 微分方程 dy/dx=(-2x)/y y=sinx/x^2,求dy y=x×2^arcsinx 求dy y=e^-x^2,求dy y=e^2x则dy? y=(tanx)/x^2,求dy y=xsin 2x,求dy y=(cos^2x) 求dy dy/dx=2y/x+3x/2y x^y-y^x=2,求dy/dx 求dy/dx=(x-y+5)/(x+y-2)