Question textLet [math: f(x,y)=ey+x2]f(x,y)={e^{y+x^2}}. What is [math: ∂f∂x(0,0)+∂f∂y(0,0)]\frac{\partial f}{\partial x}(0,0) +
\frac{\partial f}{\partial y}(0,0), the value of [math: ∂f∂x+∂f∂y]\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} at the point [math: (0,0)] ?[input] Check Question 67

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The problem asks for the value of the expression ∂f/∂x(0,0) + ∂f/∂y(0,0) given f(x,y) = e^(y + x^2).
First, consider ∂f/∂x. Since f = e^(y + x^2), the derivative with respect to x uses the chain rule: ∂f/∂x = e^(y + x^2) * ∂(y + x^2)/∂x = e^(y + x^2) * (2x).
Evaluating at (0,0): ∂f/∂x(0,0) = e^(0 + 0) * (2*0) = 1 * 0 = 0.
Next, consider ∂f/∂y. Again using the chain rule, ∂f/∂y = e^(y + x^2) * ∂(y + x^2)/∂y = e^(y + x^2) * 1 = e^(y + x^2).
Evaluating at (0,0): ∂f/∂y(0,0) = e^(0 + 0) = 1.
Now add the two partial derivatives at the point: ∂f/∂x(0,0) + ∂f/∂y(0,0) = 0 + 1 = 1.
So, the expression evaluates to 1.

If you want to double-check, plug (0,0) into the original function to confirm the derivatives behave as expected: e^(0) = 1, and the x-derivative introduces a factor of 2x which vanishes at x = 0, while the y-derivative simply reproduces e^(y + x^2) which is 1 at (0,0).

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