Assume \( y \) is constant and differentiate with respect to \( x \) to obtain \[ \begin{align*} f_x &= \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 y + 2 x + y ) \\ &= \frac{\partial}{\partial x}(x^2 y ) + \frac{\partial}{\partial x}(2 x) + \frac{\partial}{\partial x}( y ) = 2 xy + 2 + 0 = 2xy + 2 \end{align*} \] Assume \( x \) is constant and differentiate with respect to \( y \) to obtain \[ \begin{align*} f_y &= \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 y + 2 x + y ) \\ &= \frac{\partial}{\partial y}(x^2 y ) + \frac{\partial}{\partial y}(2 x) + \frac{\partial}{\partial y}( y ) = x^2 + 0 + 1 = x^2 + 1 \end{align*} \]

Differentiate with respect to \( x \) assuming \( y \) is constant \[ \begin{align*} f_x &= \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(\sin(x y) + \cos x ) \\ &= \frac{\partial}{\partial x}(\sin(x y) ) + \frac{\partial}{\partial x}(\cos x) = y \cos(x y) -\sin(x) \end{align*} \] Differentiate with respect to \( y \) assuming \( x \) is constant \[ \begin{align*} f_y &= \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(\sin(x y) + \cos x ) \\ &= \frac{\partial}{\partial y}(\sin(x y) ) + \frac{\partial}{\partial y}(\cos x) = x \cos(x y) - 0 = x \cos(x y) \end{align*} \]

Differentiate with respect to \( x \) assuming \( y \) is constant using the product rule of differentiation. \[ \begin{align*} f_x &= \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x e^{x y}) \\ &= \frac{\partial}{\partial x}(x) e^{x y} + x \frac{\partial}{\partial x}(e^{x y}) = 1 \cdot e^{x y} + x \cdot y e^{x y} = (1+xy) e^{x y} \end{align*} \] Differentiate with respect to \( y \) assuming \( x \) is constant. \[ \begin{align*} f_y &= \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x e^{x y}) \\ &= x \frac{\partial}{\partial y}(e^{x y}) = x \cdot x e^{x y} = x^2 e^{x y} \end{align*} \]

Differentiate with respect to \( x \) to obtain \[ \begin{align*} f_x &= \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(\ln(x^2+2y)) \\ &= \frac{\partial}{\partial x}(x^2+2y) \cdot \frac{1}{x^2+2y} = \frac{2x}{x^2+2y} \end{align*} \] Differentiate with respect to \( y \) \[ \begin{align*} f_y &= \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(\ln(x^2+2y)) \\ &= \frac{\partial}{\partial y}(x^2+2y) \cdot \frac{1}{x^2+2y} = \frac{2}{x^2+2y} \end{align*} \]

Find \( f_x(2 , 3) \) and \( f_y(2 , 3) \) if \( f(x , y) \) is given by \[ f(x,y) = y x^2 + 2 y \]

We first find the partial derivatives \( f_x \) and \( f_y \) \[ f_x(x,y) = 2x y \] \[ f_y(x,y) = x^2 + 2 \] We now calculate \( f_x(2 , 3) \) and \( f_y(2 , 3) \) by substituting \( x \) and \( y \) by their given values \[ f_x(2,3) = 2 (2)(3) = 12 \] \[ f_y(2,3) = 2^2 + 2 = 6 \]

1. \( f(x , y) = x e^{x + y} \)

2. \( f(x , y) = \ln ( 2 x + y x) \)

3. \( f(x , y) = x \sin(x - y) \)

2. \( f_x = 1 / x \quad \) , \( \quad f_y = 1 / (y + 2) \)

3. \( f_x = x \cos (x - y) + \sin (x - y) \quad \), \( \quad f_y = -x \cos (x - y) \)

Tables of Formulas for Derivatives

Rules of Differentiation of Functions in Calculus

Multivariable Functions

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