# 4.3 Partial derivatives  (Page 2/11)

 Page 2 / 11

Use the definition of the partial derivative as a limit to calculate $\partial f\text{/}\partial x$ and $\partial f\text{/}\partial y$ for the function

$f\left(x,y\right)=4{x}^{2}+2xy-{y}^{2}+3x-2y+5.$

$\frac{\partial f}{\partial x}=8x+2y+3,\phantom{\rule{1em}{0ex}}\frac{\partial f}{\partial y}=2x-2y-2$

The idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variable with respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a single variable. To see why this is true, first fix $y$ and define $g\left(x\right)=f\left(x,y\right)$ as a function of $x.$ Then

${g}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{g\left(x+h\right)-g\left(x\right)}{h}=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h,y\right)-f\left(x,y\right)}{h}=\frac{\partial f}{\partial x}.$

The same is true for calculating the partial derivative of $f$ with respect to $y.$ This time, fix $x$ and define $h\left(y\right)=f\left(x,y\right)$ as a function of $y.$ Then

${h}^{\prime }\left(x\right)=\underset{k\to 0}{\text{lim}}\frac{h\left(x+k\right)-h\left(x\right)}{k}=\underset{k\to 0}{\text{lim}}\frac{f\left(x,y+k\right)-f\left(x,y\right)}{k}=\frac{\partial f}{\partial y}.$

All differentiation rules from Introduction to Derivatives apply.

## Calculating partial derivatives

Calculate $\partial f\text{/}\partial x$ and $\partial f\text{/}\partial y$ for the following functions by holding the opposite variable constant then differentiating:

1. $f\left(x,y\right)={x}^{2}-3xy+2{y}^{2}-4x+5y-12$
2. $g\left(x,y\right)=\text{sin}\left({x}^{2}y-2x+4\right)$
1. To calculate $\partial f\text{/}\partial x,$ treat the variable $y$ as a constant. Then differentiate $f\left(x,y\right)$ with respect to $x$ using the sum, difference, and power rules:
$\begin{array}{cc}\hfill \frac{\partial f}{\partial x}& =\frac{\partial }{\partial x}\left[{x}^{2}-3xy+2{y}^{2}-4x+5y-12\right]\hfill \\ & =\frac{\partial }{\partial x}\left[{x}^{2}\right]-\frac{\partial }{\partial x}\left[3xy\right]+\frac{\partial }{\partial x}\left[2{y}^{2}\right]-\frac{\partial }{\partial x}\left[4x\right]+\frac{\partial }{\partial x}\left[5y\right]-\frac{\partial }{\partial x}\left[12\right]\hfill \\ & =2x-3y+0-4+0-0\hfill \\ & =2x-3y-4.\hfill \end{array}$

The derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable $x,$ so they are treated as constant terms. The derivative of the second term is equal to the coefficient of $x,$ which is $-3y.$ Calculating $\partial f\text{/}\partial y\text{:}$
$\begin{array}{cc}\hfill \frac{\partial f}{\partial y}& =\frac{\partial }{\partial y}\left[{x}^{2}-3xy+2{y}^{2}-4x+5y-12\right]\hfill \\ & =\frac{\partial }{\partial y}\left[{x}^{2}\right]-\frac{\partial }{\partial y}\left[3xy\right]+\frac{\partial }{\partial y}\left[2{y}^{2}\right]-\frac{\partial }{\partial y}\left[4x\right]+\frac{\partial }{\partial y}\left[5y\right]-\frac{\partial }{\partial y}\left[12\right]\hfill \\ & =-3x+4y-0+5-0\hfill \\ & =-3x+4y+5.\hfill \end{array}$

2. To calculate $\partial g\text{/}\partial x,$ treat the variable y as a constant. Then differentiate $g\left(x,y\right)$ with respect to x using the chain rule and power rule:
$\begin{array}{cc}\hfill \frac{\partial g}{\partial x}& =\frac{\partial }{\partial x}\left[\text{sin}\left({x}^{2}y-2x+4\right)\right]\hfill \\ & =\text{cos}\left({x}^{2}y-2x+4\right)\frac{\partial }{\partial x}\left[{x}^{2}y-2x+4\right]\hfill \\ & =\left(2xy-2\right)\text{cos}\left({x}^{2}y-2x+4\right).\hfill \end{array}$

To calculate $\partial g\text{/}\partial y,$ treat the variable $x$ as a constant. Then differentiate $g\left(x,y\right)$ with respect to $y$ using the chain rule and power rule:
$\begin{array}{cc}\hfill \frac{\partial g}{\partial y}& =\frac{\partial }{\partial y}\left[\text{sin}\left({x}^{2}y-2x+4\right)\right]\hfill \\ & =\text{cos}\left({x}^{2}y-2x+4\right)\frac{\partial }{\partial y}\left[{x}^{2}y-2x+4\right]\hfill \\ & ={x}^{2}\text{cos}\left({x}^{2}y-2x+4\right).\hfill \end{array}$

Calculate $\partial f\text{/}\partial x$ and $\partial f\text{/}\partial y$ for the function $f\left(x,y\right)=\text{tan}\left({x}^{3}-3{x}^{2}{y}^{2}+2{y}^{4}\right)$ by holding the opposite variable constant, then differentiating.

$\begin{array}{c}\frac{\partial f}{\partial x}=\left(3{x}^{2}-6x{y}^{2}\right){\text{sec}}^{2}\left({x}^{3}-3{x}^{2}{y}^{2}+2{y}^{4}\right)\hfill \\ \frac{\partial f}{\partial y}=\left(-6{x}^{2}y+8{y}^{3}\right){\text{sec}}^{2}\left({x}^{3}-3{x}^{2}{y}^{2}+2{y}^{4}\right)\hfill \end{array}$

How can we interpret these partial derivatives? Recall that the graph of a function of two variables is a surface in ${ℝ}^{3}.$ If we remove the limit from the definition of the partial derivative with respect to $x,$ the difference quotient remains:

$\frac{f\left(x+h,y\right)-f\left(x,y\right)}{h}.$

This resembles the difference quotient for the derivative of a function of one variable, except for the presence of the $y$ variable. [link] illustrates a surface described by an arbitrary function $z=f\left(x,y\right).$

In [link] , the value of $h$ is positive. If we graph $f\left(x,y\right)$ and $f\left(x+h,y\right)$ for an arbitrary point $\left(x,y\right),$ then the slope of the secant line passing through these two points is given by

$\frac{f\left(x+h,y\right)-f\left(x,y\right)}{h}.$

This line is parallel to the $x\text{-axis.}$ Therefore, the slope of the secant line represents an average rate of change of the function $f$ as we travel parallel to the $x\text{-axis.}$ As $h$ approaches zero, the slope of the secant line approaches the slope of the tangent line.

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