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) \( f(x, y)=y^{2}-y \sin (x y)+x^{3}-1 \)

To find the partial derivatives of the given functions, we will differentiate with respect to \( x \) and \( y \) while treating the other variable as a constant. For part (a): 1. To find \( \frac{\partial f}{\partial x} \): \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left((x^{2}-1)(3y+5)\right) = (2x)(3y+5) + (x^{2}-1)(0) = 2x(3y+5) = 6xy + 10x \] 2. To find \( \frac{\partial f}{\partial y} \): \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left((x^{2}-1)(3y+5)\right) = (x^{2}-1)(3) = 3(x^{2}-1) \] So the answers for part (a): - \( \frac{\partial f}{\partial x} = 6xy + 10x \) - \( \frac{\partial f}{\partial y} = 3(x^{2}-1) \) For part (b): 1. To find \( \frac{\partial f}{\partial x} \): \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(y^{2} - y \sin (xy) + x^{3} - 1\right) = 0 - y \cos(xy) \cdot y + 3x^{2} - 0 = 3x^{2} - y^{2} \cos(xy) \] 2. To find \( \frac{\partial f}{\partial y} \): \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left( y^{2} - y \sin(xy) + x^{3} - 1\right) = 2y - \left(\sin(xy) + xy \cos(xy)\right) + 0 - 0 = 2y - \sin(xy) - xy \cos(xy) \] Therefore, the results for part (b): - \( \frac{\partial f}{\partial x} = 3x^{2} - y^{2} \cos(xy) \) - \( \frac{\partial f}{\partial y} = 2y - \sin(xy) - xy \cos(xy) \)