Question

If \[ f(x,y,z)=e^{\,1-y\cos x}+yze^{\frac{-1}{1+x^2}}, \] then \[ \frac{\partial f}{\partial y} \] at \((0,1,e)\) is

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

Hint:

When evaluating a partial derivative at a point, first differentiate symbolically and then substitute the coordinates. This minimizes computational errors.

Answer 1

The Correct Option is D

Concept: To find the partial derivative of a multivariable function with respect to a variable, differentiate only with respect to that variable while treating all other variables as constants. Then substitute the given point.

Step 1: Differentiate the given function with respect to \(y\). Given, \[ f(x,y,z)=e^{\,1-y\cos x}+yze^{\frac{-1}{1+x^2}}. \] Differentiating partially with respect to \(y\), \[ \frac{\partial f}{\partial y} = -\cos x\;e^{\,1-y\cos x} + ze^{\frac{-1}{1+x^2}}. \]

Step 2: Substitute the point \((0,1,e)\). At \[ x=0,\qquad y=1,\qquad z=e, \] we have \[ \cos0=1, \] therefore, \[ \frac{\partial f}{\partial y}(0,1,e) = -e^{\,1-1} + e\cdot e^{-1}. \] Since \[ e^{0}=1 \quad\text{and}\quad e\cdot e^{-1}=1, \] we get \[ \frac{\partial f}{\partial y}(0,1,e) = -1+1=0. \] Hence, \[ \boxed{\frac{\partial f}{\partial y}(0,1,e)=0.} \] Therefore, the correct option is \[ \boxed{(D)\;0.} \]