## Definition of Partial DerivativesLet 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, using the rules and formulas of differentiation, we obtain what is called thepartial derivative of f with respect to x which is denoted by
partial derivative of f with respect to y which is denoted by
We might also use the limits to define partial derivatives of function f as follows: ## Examples with Detailed SolutionsWe now present several examples with detailed solution on how to calculate partial derivatives.
## Example 1Find the partial derivatives f_{x} and f_{y} 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 2Find f_{x} and f_{y} 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 3Find f_{x} and f_{y} if f(x , y) is given by
f(x,y) = x e^{x y}
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 4Find f_{x} and f_{y} if f(x , y) is given by
f(x,y) = \ln(x^2+2y)
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 5Find f_{x}(2 , 3) and f_{y}(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 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 valuesf _{x}(2,3) = 2 (2)(3) = 12
f _{y}(2,3) = 2^{2} + 2 = 6
## ExercisesFind partial derivatives f_{x} and f_{y} of the following functions
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) ## Answers to the Above Exercises1. f_{x} =(x + 1)e^{x + y} , f_{y} = x e^{x + y}2. f _{x} = 1 / x , f_{y} = 1 / (y + 2)
3. f _{x} = x cos (x - y) + sin (x - y) , f_{y} = -x cos (x - y)
## More References and Links to Partial Derivatives and Mtlivariable FunctionsPartial Derivative CalculatorTables of Formulas for Derivatives Rules of Differentiation of Functions in Calculus Multivariable Functions Home Page |