2 - Partial derivatives

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Calculus of Multivariable Functions

Section 2 Partial derivatives and the rules of differentiation

If a function is a multivariable function, we use the concept of partial differentiation to measure the effect of a change in one independent variable on the dependent variable, keeping the other independent variables constant. To apply the rules of calculus, at a time generally, we change only one independent variable and keep all other independent variables constant. In this way, we only look at the partial variation in the function instead of the total variation.

For example, if a function is $f(x,y)$, we use the partial differentiation with respect to $x$ , to measure the rate of change in $f(x,y)$ when only $x$ changes and $y$ remains constant. It is written as $\frac{\partial f}{\partial x}$ or simply $f_{x}$.

Similarly, we use the partial derivative with respect to $y$ (which we denote as $\frac{\partial f}{\partial y}$ or simply $f_{y}$) to measure the rate of change in $f(x,y)$ when only $y$ changes and $x$ remains constant.

To indicate that we are performing a partial differentiation and not the total differentiation, we use the sign $\frac{\partial f(x,y)}{\partial x}$ (or, simply $\frac{\partial }{\partial x}$ ) in place of $\frac{df(x)}{d(x)}$.

Partial differentiation with only one of the independent variables uses the same rules as the differentiation of one variable functions, except that while differentiating a function of several variables with respect to one independent variable, we keep all other independent variables as constants (or as coefficients). For example, consider the function $$f(x,y) = 10 + 2x + 3y$$ We compute $\frac{\partial f(x,y)}{\partial x}$ treating $\color{blue}{y}$ as a constant: $$\frac{\partial f(x,y)}{\partial x} = \frac{\partial (10)}{\partial x} + \color{red}{\frac{\partial (2x)}{\partial x}} + \frac{\partial (3\color{blue}{y})}{\partial x}$$ $$= 0 + \color{red}{2} + \color{blue}{0} = 2$$ and $\frac{\partial f(x,y)}{\partial \color{red}{y}}$ treating $\color{blue}{x}$ as a constant: $$\frac{\partial f(x,y)}{\partial y} = \frac{\partial (10)}{\partial y} + \frac{\partial (2\color{blue}{x})}{\partial y} + \color{red}{\frac{\partial (3y)}{\partial y}}$$ $$= \color{blue}{0} + 0 + \color{red}{3} = 3$$

Example: $f(x,y) = 3x^{3}y^{2}$

$\frac{\partial f(x,y)}{\partial \color{red}{x}} = 3\color{red}{(\frac{\partial }{\partial x} x^{3})} [y^{2}] = 3 \times \color{red}{3 \times x^{3 - 1}} \times y^{2} = 9 \color{red}{x^{2}} y^{2}$ (Here, $y$ is treated as a coefficient.)

$\frac{\partial f(x,y)}{\partial \color{red}{y}} = 3[x^{3}]\color{red}{(\frac{\partial }{\partial y} y^{2})} = 3 \times \color{red}{2} \times x^{3} \times \color{red}{y^{2 - 1}} = 6x^{3}\color{red}{y}$ (Here, $x$ is treated as a coefficient.)

Rules of partial differentiation:

All rules of ordinary differentiation apply, except that at a time we change only one independent variable and keep all other independent variables constant. Some of those are:
Power Rule: $f(x,y) = x^{n}y^{m}$ $$\frac{\partial f}{\partial \color{red}{x}} = \frac{\partial }{\partial \color{red}{x}} \color{red}{x^{n}}[y^{m}] = y^{m} \frac{\partial }{\partial \color{red}{x}} \color{red}{x^{n}} = \color{red}{nx^{n - 1}} y^{m}$$ $$\frac{\partial f}{\partial \color{red}{y}} = \frac{\partial }{\partial \color{red}{y}} \color{red}{y^{m}}[x^{n}] = x^{n} \frac{\partial }{\partial \color{red}{y}} \color{red}{y^{m}} = \color{red}{m}x^{n} \color{red}{y^{m - 1}}$$

Product Rule: Given $f(x,y) = \color{blue}{g(x,y)} \cdot \color{purple}{h(x,y)}$ where $\color{blue}{g}$ and $\color{purple}{h}$ are differentiable functions: $$\frac{\partial f}{\partial \color{red}{x}} = \color{blue}{g(x,y)}\frac{\color{purple}{\partial h(x,y)}}{\partial \color{red}{x}} + \color{purple}{h(x,y)}\frac{\partial \color{blue}{g(x,y)}}{\partial \color{red}{x}} \qquad \text{(Consider } y \text{ as a constant)}$$ $$\frac{\partial f}{\partial \color{red}{y}} = \color{blue}{g(x,y)}\frac{\color{purple}{\partial h(x,y)}}{\partial \color{red}{y}} + \color{purple}{h(x,y)}\frac{\partial \color{blue}{g(x,y)}}{\partial \color{red}{y}} \qquad \text{(Consider } x \text{ as a constant)}$$

Quotient Rule: Given $f(x,y) = \frac{\color{blue}{g(x,y)}}{\color{purple}{h(x,y)}}$ where $\color{blue}{g}$ and $\color{purple}{h}$ are differentiable functions and $\color{purple}{h} \neq 0$: $$\frac{\partial f}{\partial \color{red}{x}} = \frac{ \begin{bmatrix} \color{purple}{h(x,y)}\frac{\partial \color{blue}{g(x,y)}}{\partial \color{red}{x}} - \color{blue}{g(x,y)}\frac{\partial \color{purple}{h(x,y)}}{\partial \color{red}{x}} \end{bmatrix} }{(\color{purple}{h(x,y)^{2}})} \qquad \text{(Consider } y \text{ as a constant)}$$ $$\frac{\partial f}{\partial \color{red}{y}} = \frac{ \begin{bmatrix} \color{purple}{h(x,y)}\frac{\partial \color{blue}{g(x,y)}}{\partial \color{red}{y}} - \color{blue}{g(x,y)}\frac{\partial \color{purple}{h(x,y)}}{\partial \color{red}{y}} \end{bmatrix} }{(\color{purple}{h(x,y)^{2}})} \qquad \text{(Consider } x \text{ as a constant)}$$

Steps at a glance while calculating the partial derivative with respect to $\color{red}{x}$:

Step 1: Use $\frac{\partial f(x,y)}{\partial x}$ in place of $\frac{df(x,y)}{dx}$

Step 2: Consider $y$ (and all other variables except $x$) as the constant and/or coefficient

Step 3: Apply the rules of ordinary differentiation

Example 1: Find the first-order partial derivatives for each of the following functions:

$f(x,y) = 10 + 2x^{2}+y^{3}$ $$\frac{\partial f(x,y)}{\partial x} = 0 + 4x + 0 = 4x$$ $$\frac{\partial f(x,y)}{\partial y} = 0 + 0 + 3y^{2} = 3y^{2}$$
$f(x,y) = 10 + 2x^{2}y^{3}$ $$\frac{\partial f(x,y)}{\partial x} = 0 + 4xy^{3} = 4xy^{3}$$ $$\frac{\partial f(x,y)}{\partial y} = 0 + 6x^{2}y^{2} = 6x^{2}y^{2}$$
$f(x,y) = 2x^{0.5}y^{0.5}$ $$\frac{\partial f(x,y)}{\partial x} = 2\color{red}{(0.5)x^{0.5 - 1}}y^{0.5} = \color{red}{x^{-0.5}}y^{0.5} = \frac{y^{0.5}}{\color{red}{x^{0.5}}}$$ $$\frac{\partial f(x,y)}{\partial y} = \frac{x^{0.5}}{y^{0.5}}$$
$f(x,y) = x^{0.6}y^{0.3}$ $$\frac{\partial f(x,y)}{\partial x} = \color{red}{0.6x^{0.6 - 1}}y^{0.3} = \color{red}{0.6x^{-0.4}}y^{0.3} = \color{red}{0.6}\frac{y^{0.3}}{\color{red}{x^{0.4}}}$$ $$\frac{\partial f(x,y)}{\partial y} = 0.3\frac{x^{0.6}}{y^{0.7}}$$
$f(x,y,z) = 10x + 12y + 20z$ $$f_{x} = 10, \qquad f_{y} = 12, \qquad f_{z} = 20$$

Example 2: Find $f_{x} = \frac{\partial f(x,y)}{\partial x}$ and $f_{y} = \frac{\partial f(x,y)}{\partial y}$:

$f(x,y) = \color{blue}{2x^{2}}\color{purple}{(4x + 6y)}$
Using the product rule: $$f_{x} = \color{blue}{2x^{2}}\color{purple}{(4)} + \color{purple}{(4x + 6y)}(\color{blue}{4x}) = 24x^{2} + 24xy$$ $$f_{y} = \color{blue}{2x^{2}}\color{purple}{(6)} + \color{purple}{(4x + 6y)}(\color{blue}{0}) = 12x^{2}$$
$f(x,y) = \frac{\color{blue}{x + y}}{\color{purple}{2y}}$
Using the quotient rule: $$f_{x} = \frac{\begin{bmatrix} (\color{purple}{2y})(\color{blue}{1}) - (\color{blue}{x + y})(\color{purple}{0}) \end{bmatrix}}{\color{purple}{4y^{2}}} = \frac{2y}{4y^{2}} = \frac{1}{2y}$$ $$f_{y} = \frac{\begin{bmatrix} (\color{purple}{2y})(\color{blue}{1}) - (\color{blue}{x + y})(\color{purple}{2}) \end{bmatrix}}{\color{purple}{4y^{2}}} = \frac{-2x}{4y^{2}} = -\frac{x}{2y}$$

Section 2

Partial derivatives and the rules of differentiation

Rules of partial differentiation:

Steps at a glance while calculating the partial derivative with respect to \(\color{red}{x}\):

You are recommended to do more practice from a suggested book to build up your confidence.