f(x,y)= $\left\{\begin{matrix} 1-cos(\frac{x^2}{y^2}) \ \sqrt{x^2+y^2} \end{matrix}\right.$ ;if y≠0 & n∈R, then

Question

f(x,y)= $\left\{\begin{matrix} 1-cos(\frac{x^2}{y^2}) \  \sqrt{x^2+y^2} \end{matrix}\right.$  ;if y≠0 & n∈R, then

cdquestions Admin · Accepted Answer

The correct option is (B,C) :   f is continuous at (0,0) but f is not differentiable at (0,0). Partial derivatives of f exist at (0,0) of equal to o.

Answer

Directional derivatives of f exist at (0,0) & equal to 0.

Answer

f is continuous at (0,0) but f is not differentiable at (0,0).

Answer

Partial derivatives of f exist at (0,0) of equal to o.

Answer

f is differentiable at (0,0).

f(x,y)=\(\left\{\begin{matrix} 1-cos(\frac{x^2}{y^2}) \\ \sqrt{x^2+y^2} \end{matrix}\right.\) ;if y≠0 & n∈R, then