Partial Derivatives in Calculus

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
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