Question

f(x,y) = y^3 + ysin2x + e^(x+y) is diffrentiable at (1,-1).

Answer 1

$f(x,y) = y^3 + ysin2x + e^{x+y}$;

$f(x,y)$ is differential at (a,b) if f is continuous at (a,b) and the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are both defined and continuous at (a,b).

f is continuous as a sum of continuous functions.

Let's find partial derivatives of f.

$\frac{\partial f}{\partial x}=2ycos2x+e^{x+y}.$

$\frac{\partial f}{\partial y}=2y^2+sin2x+e^{x+y}.$

The partial derivatives are both defined at (-1,1) and continuous as the sums of continuous functions.

Hence $f(x,y)=y^3+ysin2x+e^{x+y}$ is differettiable at (-1; 1)