Question

If \[ f(x,y,z,w)=x^{2}e^{2y+3z}\cos(4w), \] then \[ \frac{\partial f}{\partial z} \] at \((2,0,-2,1)\) is

• A. \[ \frac{8\cos4}{e^{4}} \]
• B. \[ \frac{12\cos4}{e^{6}} \]
• C. \[ \cos4 \]
• D. \[ e \]

Hint:

For functions of the form \[ f=x^m e^{ax+by+cz}\sin(kz) \quad\text{or}\quad f=x^m e^{ax+by+cz}\cos(kz), \] while taking the partial derivative with respect to one variable:
• Treat all other variables as constants.
• Apply the chain rule only to the exponential term.
• Substitute the given point only after obtaining the derivative. This saves time and reduces calculation errors in competitive examinations.

Answer 1

The Correct Option is B

Concept: A partial derivative measures the rate of change of a multivariable function with respect to one variable while treating all the remaining variables as constants. For an exponential function, \[ \frac{\partial}{\partial z}\left(e^{u}\right) = e^{u}\cdot\frac{\partial u}{\partial z}, \] which follows from the chain rule. Also, \[ \frac{\partial}{\partial z}\left(x^2\right)=0,\qquad \frac{\partial}{\partial z}\left(\cos4w\right)=0, \] since \(x\) and \(w\) are treated as constants while differentiating with respect to \(z\). Hence, only the exponential term contributes to the derivative.

Step 1: Write the given function.
The given function is \[ f(x,y,z,w) = x^{2}e^{2y+3z}\cos(4w). \] We have to calculate \[ \frac{\partial f}{\partial z}. \]

Step 2: Differentiate with respect to \(z\).
Since \(x^2\) and \(\cos(4w)\) are constants with respect to \(z\), \[ \frac{\partial f}{\partial z} = x^{2}\cos(4w) \cdot \frac{\partial}{\partial z} \left(e^{2y+3z}\right). \] Applying the chain rule, \[ \frac{\partial}{\partial z} \left(e^{2y+3z}\right) = e^{2y+3z}\cdot \frac{\partial}{\partial z}(2y+3z). \] Since \[ \frac{\partial}{\partial z}(2y+3z)=3, \] therefore, \[ \frac{\partial}{\partial z} \left(e^{2y+3z}\right) = 3e^{2y+3z}. \] Hence, \[ \boxed{ \frac{\partial f}{\partial z} = 3x^{2}e^{2y+3z}\cos(4w) }. \]

Step 3: Substitute the given point \((2,0,-2,1)\).
Substituting, \[ x=2,\qquad y=0,\qquad z=-2,\qquad w=1. \] Therefore, \[ \frac{\partial f}{\partial z} = 3(2)^2 e^{2(0)+3(-2)} \cos(4). \]

Step 4: Simplify the expression.
First, \[ (2)^2=4. \] Also, \[ 2(0)+3(-2) = -6. \] Hence, \[ \frac{\partial f}{\partial z} = 3\times4\times e^{-6}\cos4. \] Therefore, \[ = 12e^{-6}\cos4. \] Since \[ e^{-6}=\frac1{e^6}, \] we obtain \[ \boxed{ \frac{\partial f}{\partial z} = \frac{12\cos4}{e^6} }. \]

Step 5: Compare with the given options.
The obtained result is \[ \frac{12\cos4}{e^6}, \] which exactly matches \[ \boxed{\text{Option (B)}}. \] Hence, \[ \boxed{\text{Correct Answer = (2)}} \]

Step 6: Alternative shortcut.
Instead of differentiating the entire function, remember that \[ \frac{\partial}{\partial z} \left(e^{2y+3z}\right) = 3e^{2y+3z}. \] Thus, \[ \frac{\partial f}{\partial z} = 3f(x,y,z,w). \] Now simply evaluate \[ f(2,0,-2,1) = 4e^{-6}\cos4. \] Therefore, \[ \frac{\partial f}{\partial z} = 3\left(4e^{-6}\cos4\right) = \frac{12\cos4}{e^6}. \] This shortcut is very useful in objective examinations.