# Partial Derivatives

 Let f(x,y) be a function with two variables. If we keep y constant and differentiate f (assuming f is differentiable) with respect to the variable x, we obtain what is called the partial derivative of f with respect to x which is denoted by \frac{\partial f}{\partial x} \;\; \text{or} \; \; f_x Similarly If we keep x constant and differentiate f (assuming f is differentiable) with respect to the variable y, we obtain what is called the partial derivative of f with respect to y which is denoted by \frac{\partial f}{\partial y} \;\; \text{or} \; \; f_y We might also use the limits to define partial derivatives of function f as follows: \dfrac{\partial f}{\partial x} = \lim_{h\to 0} \dfrac{f(x+h,y)-f(x,y)}{h} \dfrac{\partial f}{\partial y} = \lim_{k\to 0} \dfrac{f(x,y+k)-f(x,y)}{k} We now present several examples with detailed solution on how to calculate partial derivatives. Example 1: Find the partial derivatives fx and fy if f(x , y) is given by f(x,y) = x^2 y + 2 x + y Solution to Example 1: Assume y is constant and differentiate with respect to x to obtain 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 Now assume x is constant and differentiate with respect to y to obtain 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 Example 2: Find fx and fy if f(x , y) is given by f(x,y) = \sin(x y) + \cos x Solution to Example 2: Differentiate with respect to x assuming y is constant 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) Differentiate with respect to y assuming x is constant 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) Example 3: Find fx and fy if f(x , y) is given by f(x,y) = x e^{x y} Solution to Example 3: Differentiate with respect to x assuming y is constant using the product rule of differentiation. 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} Differentiate with respect to y assuming x is constant. 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} Example 4: Find fx and fy if f(x , y) is given by f(x,y) = \ln(x^2+2y) Solution to Example 4: Differentiate with respect to x to obtain f_x = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(\ln(x^2+2y)) \\\\ = \frac{\partial}{\partial x}(x^2+2y) \cdot \dfrac{1}{x^2+2y} = \dfrac{2x}{x^2+2y} Differentiate with respect to y f_y = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(\ln(x^2+2y)) \\\\ = \frac{\partial}{\partial y}(x^2+2y) \cdot \dfrac{1}{x^2+2y} = \dfrac{2}{x^2+2y} Example 5: Find fx(2 , 3) and fy(2 , 3) if f(x , y) is given by f(x,y) = y x^2 + 2 y Solution to Example 5: We first find the partial derivatives fx and fy fx(x,y) = 2x y fy(x,y) = x2 + 2 We now calculate fx(2 , 3) and fy(2 , 3) by substituting x and y by their given values fx(2,3) = 2 (2)(3) = 12 fy(2,3) = 22 + 2 = 6 Exercise: Find partial derivatives fx and fy of the following functions 1. f(x , y) = x ex + y 2. f(x , y) = ln ( 2 x + y x) 3. f(x , y) = x sin(x - y) Answer to Above Exercise: 1. fx =(x + 1)ex + y , fy = x ex + y 2. fx = 1 / x , fy = 1 / (y + 2) 3. fx = x cos (x - y) + sin (x - y) , fy = -x cos (x - y) More on partial derivatives and mutlivariable functions. Multivariable Functions Home Page