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
\frac{\partial f}{\partial x} \;\; \text{or} \; \; f_x
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.
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
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)
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 4: Find 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}
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 values
f _{x}(2,3) = 2 (2)(3) = 12
f _{y}(2,3) = 2^{2} + 2 = 6
Exercise: Find 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) Answer to Above Exercise:
1. 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)
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