Question

For the function \(f(x,y)=e^{x\cos(y)\), what is the value of \(\dfrac{\partial^{2}f}{\partial x\,\partial y}\) at \((x=0,y=\pi/2)\)?}

• A. 0
• B. 1
• C. \(-1\)
• D. \(e^{\pi/2}\)

Hint:

For mixed partials of separable forms like \(e^x\cos y\), differentiate one factor at a time; evaluation at special angles often simplifies to \(0,\pm1\).

Answer 1

The Correct Option is C

Step 1: First differentiate w.r.t \(y\): \(f_y=\frac{\partial}{\partial y}(e^{x}\cos y)=-e^{x}\sin y\).
Step 2: Differentiate w.r.t \(x\): \(f_{xy}=\frac{\partial}{\partial x}(-e^{x}\sin y)=-e^{x}\sin y\).
Step 3: Evaluate at \((0,\pi/2)\): \(f_{xy}(0,\pi/2)=-e^{0}\sin(\pi/2)=-1\).
Thus the value is \(\boxed{-1}\).