Question

Let \( f(x, y) = \sqrt{x^3 y} \sin \left( \dfrac{\pi}{2} e^{(\frac{y}{x} - 1)} \right) + xy \cos \left( \frac{\pi}{3} e^{(\frac{x}{y} - 1)} \right) \) for \( (x, y) \in \mathbb{R}^2, x > 0, y > 0 \). 
Then \( f_x(1, 1) + f_y(1, 1) = \) .......... 
 

Hint:

When evaluating partial derivatives, always simplify the expression and evaluate the result at the given point.

Answer 1

Correct Answer: 3

Step 1: Compute \( f_x(x, y) \)

We begin with the function: \( f(x, y) = \sqrt{x^3 y} \sin\left( \frac{\pi}{2} e^{(\frac{y}{x} - 1)} \right) + xy \cos\left( \frac{\pi}{3} e^{(\frac{x}{y} - 1)} \right) \).

• Differentiate \( \sqrt{x^3 y} \) with respect to \( x \): \(\frac{\partial}{\partial x}\left(\sqrt{x^3 y}\right) = \frac{3}{2}\sqrt{\frac{y}{x^3}}\).
• Derivative of the sine term : \(\cos\left( \frac{\pi}{2} e^{(\frac{y}{x} - 1)} \right) \cdot \left(\frac{-\pi y e^{(\frac{y}{x} - 1)}}{2x^2}\right)\).

The partial derivative \( f_x(x, y) \) is:

\( \frac{3}{2}\sqrt{\frac{y}{x^3}} \sin\left( \frac{\pi}{2} e^{(\frac{y}{x} - 1)} \right) + \sqrt{x^3 y} \cdot \cos\left( \frac{\pi}{2} e^{(\frac{y}{x} - 1)} \right) \cdot \left(\frac{-\pi y e^{(\frac{y}{x} - 1)}}{2x^2}\right) + y \cos\left( \frac{\pi}{3} e^{(\frac{x}{y} - 1)} \right) + x \left(-\sin\left( \frac{\pi}{3} e^{(\frac{x}{y} - 1)} \right) \cdot \left(\frac{\pi e^{(\frac{x}{y} - 1)}}{3y}\right)\right) \).

Step 2: Compute \( f_y(x, y) \)

• Differentiate \( \sqrt{x^3 y} \) with respect to \( y \): \(\frac{\partial}{\partial y}\left(\sqrt{x^3 y}\right) = \frac{1}{2}\sqrt{\frac{x^3}{y}}\).
• Derivative of the sine term : \(\cos\left( \frac{\pi}{2} e^{(\frac{y}{x} - 1)} \right) \cdot \left(\frac{\pi e^{(\frac{y}{x} - 1)}}{2x}\right)\).

The partial derivative \( f_y(x, y) \) is:

\( \frac{1}{2}\sqrt{\frac{x^3}{y}} \sin\left( \frac{\pi}{2} e^{(\frac{y}{x} - 1)} \right) + \sqrt{x^3 y} \cdot \cos\left( \frac{\pi}{2} e^{(\frac{y}{x} - 1)} \right) \cdot \left(\frac{\pi e^{(\frac{y}{x} - 1)}}{2x}\right) + x \cos\left( \frac{\pi}{3} e^{(\frac{x}{y} - 1)} \right) + y \left(-\sin\left( \frac{\pi}{3} e^{(\frac{x}{y} - 1)} \right) \cdot \left(\frac{-\pi x e^{(\frac{x}{y} - 1)}}{3y^2}\right)\right) \).

Step 3: Evaluate at \( (1, 1) \)

Both \( e^{(\frac{y}{x} - 1)} \) and \( e^{(\frac{x}{y} - 1)} \) evaluate to \( e^0 = 1 \) when \( (x, y) = (1, 1) \). Therefore:

• Sine and cosine arguments are multiples of \(\pi\), yielding easy evaluations such as \( \sin(\frac{\pi}{2}) = 1 \) and \( \cos(\frac{\pi}{3}) = \frac{1}{2} \).

Thus, compute:

\( f_x(1, 1) = \frac{3}{2} \cdot 1 + 1 \cdot \cos(\frac{\pi}{2}) \cdot 0 + 1 \cdot \frac{1}{2} - 1 \cdot 0 = 2\).

\( f_y(1, 1) = \frac{1}{2} \cdot 1 + 1 \cdot \cos(\frac{\pi}{2}) \cdot 0 + 1 \cdot \frac{1}{2} - 1 \cdot 0 = 1\).

Final Calculation

\( f_x(1, 1) + f_y(1, 1) = 2 + 1 = 3 \).