2018年5月28日 · How to find $\lim _{x,y \to 0}x^2y^2 \ln (x^2+y^2)$? Could $x^2y^2 \ln (x^2+y^2)=\frac{\ln(x^2+y^2)}{x^2+y^2}x^2y^2(x^2+y^2)$ be the way?

2015年4月13日 · I want to calculate limit of $\lim_{(x,y) \rightarrow(0,0)}y \ln (x^2+y^2)$. How to do that? From iterated limits i know that limit exists for certain, but how to show that it is equal to zero then?

设z＝ln（x^2+y^2）求偏导数，如图，第十题az/ax=2x/（x^2+y^2)，a^2z/ax^2=2（-x^2+y^2)/ [（x^2+y^2)]的平方。 偏导数是通过极限来定义的，按定义写出某点 (x0,y0)处偏导数的极限表 …

2013年7月10日 · ln[根号（x^2+y^2）] =arctany/x 求dy解：1/2*ln(x^2+y^2)=arctany/x两边对x求导，得1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2化简得y'=(x+y)/(x-y)

2017年9月15日 · How do you find the derivative of y = ln(x2 + y2)? We have: y = ln(x2 + y2) Method 1: Implicit differentiation, as is: Using the chain rule: ∴ (1 − 2y x2 +y2) dy dx = 2x x2 +y2. ∴ (x2 + y2 −2y) dy dx = 2x. ∴ dy dx = 2x x2 + y2 −2y. Method 2: Use exponentials: y = ln(x2 + y2) ⇔ ey = x2 +y2. Implicit differentiation yields:

2010年12月20日 · 当（x,y)趋于(0,0)时，(x+2y)ln(x^2+y^2)的极限怎么求如图，最后一步：无穷小量×有界量 还是无穷小量

\lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx \sum_{n=0}^{\infty}\frac{3}{2^n} Show More

Find the maxima and the minima by solving ∇f = 0, then look for extreme values on the axis x = 0, y = 0.

If we have: f(x, y) = \ln(x^2 + y^2) Note that the y term is missing here, and using the chain rule with \ln, we get: d/dx ~(\ln(x^2 + y^2)) = \dfrac{1}{x^2 + y^2}~d/dx~ (x^2 + y^2) = \dfrac{1}{x^2 + y^2}(2x+0) =\dfrac{2x}{x^2 + y^2} ...

那么ln(1+x^2y^)等价于x^2y^2， sin(x^2y^2)等价于x^2y^2 所以得到 原极限 =lim(x,y->0) x^2y^2 /x^2y^2 =1 故极限值为1