Q2: Find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) of the following functions: (a) \( f(x, y)=\left(x^{2}-1\right)(3 y+5) \) (b) \( f(x, y)=y^{2}-y \sin (x y)+x^{3}-1 \)

Alright, I need to find the partial derivatives of the given functions with respect to \( x \) and \( y \). Let's tackle them one by one. **Function (a):** \( f(x, y) = (x^{2} - 1)(3y + 5) \) First, I'll expand the function to simplify the differentiation process. Multiplying the terms inside the parentheses: \[ f(x, y) = (x^{2} - 1)(3y + 5) = 3x^{2}y + 5x^{2} - 3y - 5 \] Now, I'll find the partial derivatives. 1. **Partial derivative with respect to \( x \):** \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(3x^{2}y) + \frac{\partial}{\partial x}(5x^{2}) + \frac{\partial}{\partial x}(-3y) + \frac{\partial}{\partial x}(-5) \] Calculating each term: - \( \frac{\partial}{\partial x}(3x^{2}y) = 6xy \) - \( \frac{\partial}{\partial x}(5x^{2}) = 10x \) - \( \frac{\partial}{\partial x}(-3y) = 0 \) (since \( y \) is treated as a constant) - \( \frac{\partial}{\partial x}(-5) = 0 \) (constant term) So, \[ \frac{\partial f}{\partial x} = 6xy + 10x \] 2. **Partial derivative with respect to \( y \):** \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(3x^{2}y) + \frac{\partial}{\partial y}(5x^{2}) + \frac{\partial}{\partial y}(-3y) + \frac{\partial}{\partial y}(-5) \] Calculating each term: - \( \frac{\partial}{\partial y}(3x^{2}y) = 3x^{2} \) - \( \frac{\partial}{\partial y}(5x^{2}) = 0 \) (since \( x^{2} \) is treated as a constant) - \( \frac{\partial}{\partial y}(-3y) = -3 \) - \( \frac{\partial}{\partial y}(-5) = 0 \) (constant term) So, \[ \frac{\partial f}{\partial y} = 3x^{2} - 3 \] **Function (b):** \( f(x, y) = y^{2} - y \sin(xy) + x^{3} - 1 \) This function involves a product of \( y \) and \( \sin(xy) \), so I'll need to use the product rule for differentiation. 1. **Partial derivative with respect to \( x \):** \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(y^{2}) - \frac{\partial}{\partial x}(y \sin(xy)) + \frac{\partial}{\partial x}(x^{3}) - \frac{\partial}{\partial x}(1) \] Calculating each term: - \( \frac{\partial}{\partial x}(y^{2}) = 0 \) (since \( y^{2} \) is treated as a constant) - \( \frac{\partial}{\partial x}(y \sin(xy)) = y \cos(xy) \cdot \frac{\partial}{\partial x}(xy) = y \cos(xy) \cdot y = y^{2} \cos(xy) \) - \( \frac{\partial}{\partial x}(x^{3}) = 3x^{2} \) - \( \frac{\partial}{\partial x}(1) = 0 \) (constant term) So, \[ \frac{\partial f}{\partial x} = -y^{2} \cos(xy) + 3x^{2} \] 2. **Partial derivative with respect to \( y \):** \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(y^{2}) - \frac{\partial}{\partial y}(y \sin(xy)) + \frac{\partial}{\partial y}(x^{3}) - \frac{\partial}{\partial y}(1) \] Calculating each term: - \( \frac{\partial}{\partial y}(y^{2}) = 2y \) - \( \frac{\partial}{\partial y}(y \sin(xy)) = \sin(xy) + y \cos(xy) \cdot \frac{\partial}{\partial y}(xy) = \sin(xy) + y \cos(xy) \cdot x = \sin(xy) + xy \cos(xy) \) - \( \frac{\partial}{\partial y}(x^{3}) = 0 \) (since \( x^{3} \) is treated as a constant) - \( \frac{\partial}{\partial y}(1) = 0 \) (constant term) So, \[ \frac{\partial f}{\partial y} = 2y - \sin(xy) - xy \cos(xy) \] **Summary:** - For function (a): - \( \frac{\partial f}{\partial x} = 6xy + 10x \) - \( \frac{\partial f}{\partial y} = 3x^{2} - 3 \) - For function (b): - \( \frac{\partial f}{\partial x} = -y^{2} \cos(xy) + 3x^{2} \) - \( \frac{\partial f}{\partial y} = 2y - \sin(xy) - xy \cos(xy) \)