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
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
Updated On: Aug 18, 2024
- Directional derivatives of f exist at (0,0) & equal to 0.
- 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.
- f is differentiable at (0,0).
Show Solution
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The Correct Option is B, C
Solution and Explanation
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.
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.
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