Again These Work Just Like Real valued Functions if X and Y Are Twice Differentiable Fuctions of T

Learning Objectives

• four.v.1 State the chain rules for i or two independent variables.
• 4.5.two Employ tree diagrams equally an aid to agreement the chain rule for several contained and intermediate variables.
• 4.5.3 Perform implicit differentiation of a part of ii or more variables.

In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of ii functions. The same matter is truthful for multivariable calculus, but this fourth dimension nosotros take to deal with more than one form of the chain rule. In this section, we report extensions of the chain rule and acquire how to take derivatives of compositions of functions of more i variable.

Concatenation Rules for One or Two Independent Variables

Recall that the concatenation rule for the derivative of a composite of two functions can exist written in the form

$$\frac{d}{d\mathrm{10}}\left(f\left(\mathrm{chiliad}\left(x\right)\right)\right)=f\prime \left(g\left(x\right)\right)\mathrm{thousand}\prime \left(x\right).$$

In this equation, both $f\left(x\right)$ and $g\left(\mathrm{10}\right)$ are functions of i variable. At present suppose that $f$ is a function of 2 variables and $\mathrm{one\; thousand}$ is a function of one variable. Or perhaps they are both functions of two variables, or even more. How would we summate the derivative in these cases? The post-obit theorem gives us the answer for the example of one contained variable.

Theorem iv.viii

Chain Rule for One Independent Variable

Suppose that $x=g\left(t\right)$ and $y=h\left(t\right)$ are differentiable functions of $t$ and $z=f\left(x,y\right)$ is a differentiable function of $x\text{and}y.$ Then $z=f\left(x\left(t\right),y\left(t\right)\right)$ is a differentiable function of $t$ and

$$\frac{dz}{dt}=\frac{\partial z}{\partial x}·\frac{d\mathrm{ten}}{dt}+\frac{\partial z}{\partial y}·\frac{dy}{dt},$$

(four.29)

where the ordinary derivatives are evaluated at $t$ and the fractional derivatives are evaluated at $\left(x,y\right).$

Proof

The proof of this theorem uses the definition of differentiability of a function of two variables. Suppose that f is differentiable at the point $P\left(x_0,y_0\right),$ where ${\mathrm{ten}}_0=k\left(t_0\right)$ and $y_0=h\left(t_0\right)$ for a fixed value of $t_0.$ We wish to prove that $z=f\left(x\left(t\right),y\left(t\right)\right)$ is differentiable at $t=t_0$ and that Equation 4.29 holds at that point as well.

Since $f$ is differentiable at $P,$ we know that

$$z\left(t\right)=f\left(\mathrm{10},y\right)=f\left({\mathrm{ten}}_0,y_0\right)+f_x\left(x_0,y_0\right)\left(x-{\mathrm{10}}_0\right)+f_y\left({\mathrm{ten}}_0,y_0\right)\left(y-y_0\right)+\mathrm{East}\left(x,y\right),$$

(4.30)

where $\underset{\left(\mathrm{10},y\right)\to \left(x_0,y_0\right)}{\text{lim}}\frac{E\left(\mathrm{ten},y\right)}{\sqrt{{\left(x-{\mathrm{ten}}_0\right)}^2+{\left(y-y_0\right)}^2}}=0.$ We so subtract $z_0=f\left({\mathrm{ten}}_0,y_0\right)$ from both sides of this equation:

$$\begin{array}{cc}\hfill z\left(t\right)-z\left(t_0\right)& =f\left(\mathrm{ten}\left(t\right),y\left(t\right)\right)-f\left(x\left(t_0\right),y\left(t_0\right)\right)\hfill \\ & =f_x\left({\mathrm{10}}_0,y_0\right)\left(x\left(t\right)-\mathrm{ten}\left(t_0\right)\right)+f_y\left(x_0,y_0\right)\left(y\left(t\right)-y\left(t_0\right)\right)+E\left(x\left(t\right),y\left(t\right)\right).\hfill \end{array}$$

Side by side, we divide both sides by $t-t_0\text{:}$

$$\frac{z\left(t\right)-z\left(t_0\right)}{t-t_0}=f_x\left(x_0,y_0\right)\left(\frac{x\left(t\right)-x\left(t_0\right)}{t-t_0}\right)+f_y\left(x_0,y_0\right)\left(\frac{y\left(t\right)-y\left(t_0\right)}{t-t_0}\right)+\frac{E\left(x\left(t\right),y\left(t\right)\right)}{t-t_0}.$$

Then we have the limit every bit $t$ approaches $t_0\text{:}$

$$\begin{array}{cc}\hfill \underset{t\to t_0}{\text{lim}}\frac{z\left(t\right)-z\left(t_0\right)}{t-t_0}& =f_x\left(x_0,y_0\right)\underset{t\to t_0}{\text{lim}}\left(\frac{\mathrm{ten}\left(t\right)-x\left(t_0\right)}{t-t_0}\right)+f_y\left(x_0,y_0\right)\underset{t\to t_0}{\text{lim}}\left(\frac{y\left(t\right)-y\left(t_0\right)}{t-t_0}\right)\hfill \\ & +\underset{t\to t_0}{\text{lim}}\frac{E\left(x\left(t\right),y\left(t\right)\right)}{t-t_0}.\hfill \end{array}$$

The left-hand side of this equation is equal to $dz\text{/}dt,$ which leads to

$$\frac{dz}{dt}=f_x\left(x_0,y_0\right)\frac{dx}{dt}+f_y\left(x_0,y_0\right)\frac{dy}{dt}+\underset{t\to t_0}{\text{lim}}\frac{E\left(x\left(t\right),y\left(t\right)\right)}{t-t_0}.$$

The final term can exist rewritten as

$$\begin{array}{cc}\hfill \underset{t\to t_0}{\text{lim}}\frac{\mathrm{East}\left(\mathrm{10}\left(t\right),y\left(t\right)\right)}{t-t_0}& =\underset{t\to t_0}{\text{lim}}\left(\frac{\mathrm{Due\; east}\left(\mathrm{10},y\right)}{\sqrt{{\left(x-{\mathrm{ten}}_0\right)}^{\mathrm{ii}}+{\left(y-y_0\right)}^2}}\frac{\sqrt{{\left(\mathrm{10}-{\mathrm{ten}}_0\right)}^2+{\left(y-y_0\right)}^2}}{t-t_0}\right)\hfill \\ & =\underset{t\to t_0}{\text{lim}}\left(\frac{\mathrm{Due\; east}\left(x,y\right)}{\sqrt{{\left(\mathrm{ten}-x_0\right)}^2+{\left(y-y_0\right)}^2}}\right)\underset{t\to t_0}{\text{lim}}\left(\frac{\sqrt{{\left(\mathrm{ten}-x_0\right)}^2+{\left(y-y_0\right)}^2}}{t-t_0}\right).\hfill \end{array}$$

As $t$ approaches $t_0,$ $\left(x\left(t\right),y\left(t\right)\right)$ approaches $\left(x\left(t_0\right),y\left(t_0\right)\right),$ and so nosotros can rewrite the last product every bit

$$\underset{\left(\mathrm{10},y\right)\to \left(x_0,y_0\right)}{\text{lim}}\left(\frac{E\left(\mathrm{ten},y\right)}{\sqrt{{\left(\mathrm{ten}-{\mathrm{10}}_0\right)}^2+{\left(y-y_0\right)}^{\mathrm{ii}}}}\right)\underset{\left(\mathrm{10},y\right)\to \left(x_0,y_0\right)}{\text{lim}}\left(\frac{\sqrt{{\left(x-x_0\right)}^{\mathrm{two}}+{\left(y-y_0\right)}^2}}{t-t_0}\right).$$

Since the first limit is equal to zippo, nosotros need simply show that the second limit is finite:

$$\begin{array}{cc}\hfill \underset{\left(x,y\right)\to \left(x_0,y_0\right)}{\text{lim}}\left(\frac{\sqrt{{\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2}}{t-t_0}\right)& =\underset{\left(\mathrm{ten},y\right)\to \left(x_0,y_0\right)}{\text{lim}}\left(\sqrt{\frac{{\left(\mathrm{ten}-{\mathrm{10}}_0\right)}^2+{\left(y-y_0\right)}^{\mathrm{ii}}}{{\left(t-t_0\right)}^2}}\right)\hfill \\ & =\underset{\left(x,y\right)\to \left(x_0,y_0\right)}{\text{lim}}\left(\sqrt{{\left(\frac{\mathrm{10}-{\mathrm{ten}}_0}{t-t_0}\right)}^2+{\left(\frac{y-y_0}{t-t_0}\right)}^2}\right)\hfill \\ & =\sqrt{{\left(\underset{\left(x,y\right)\to \left({\mathrm{ten}}_0,y_0\right)}{\text{lim}}\left(\frac{x-{\mathrm{ten}}_0}{t-t_0}\right)\right)}^2+{\left(\underset{\left(x,y\right)\to \left(x_0,y_0\right)}{\text{lim}}\left(\frac{y-y_0}{t-t_0}\right)\right)}^2}.\hfill \end{array}$$

Since $\mathrm{10}\left(t\right)$ and $y\left(t\right)$ are both differentiable functions of $t,$ both limits inside the concluding radical exist. Therefore, this value is finite. This proves the chain rule at $t=t_0;$ the rest of the theorem follows from the assumption that all functions are differentiable over their entire domains.

□

Closer examination of Equation 4.29 reveals an interesting pattern. The starting time term in the equation is $\frac{\partial f}{\partial x}·\frac{dx}{dt}$ and the 2d term is $\frac{\partial f}{\partial y}·\frac{dy}{dt}.$ Recall that when multiplying fractions, cancelation can be used. If we treat these derivatives as fractions, and so each product "simplifies" to something resembling $\partial f\text{/}dt.$ The variables $\mathrm{ten}\text{and}y$ that disappear in this simplification are often called intermediate variables: they are independent variables for the function $f,$ only are dependent variables for the variable $t.$ Ii terms appear on the right-paw side of the formula, and $f$ is a function of 2 variables. This pattern works with functions of more than 2 variables as well, equally nosotros see later in this section.

Example 4.26

Using the Chain Dominion

Calculate $dz\text{/}dt$ for each of the following functions:

1. $z=f\left(x,y\right)=4x^2+\mathrm{three}{y}^2,\mathrm{ten}=x\left(t\right)=\text{sin}t,y=y\left(t\right)=\text{cos}t$
2. $z=f\left(x,y\right)=\sqrt{{\mathrm{ten}}^2-y^2},x=x\left(t\right)=e^{2t},y=y\left(t\right)=e^{\text{−}t}$

Checkpoint iv.23

Calculate $dz\text{/}dt$ given the following functions. Express the concluding respond in terms of $t.$

$$z=f\left(x,y\right)=x^2-3xy+2y^2,\mathrm{10}=x\left(t\right)=\mathrm{three}\text{sin}\mathrm{two}t,y=y\left(t\right)=4\text{cos}2t$$

It is often useful to create a visual representation of Equation 4.29 for the concatenation rule. This is called a tree diagram for the chain rule for functions of 1 variable and information technology provides a way to recall the formula (Figure 4.34). This diagram tin can be expanded for functions of more than one variable, as we shall meet very shortly.

Figure iv.34 Tree diagram for the case $\frac{dz}{dt}=\frac{\partial z}{\partial x}·\frac{dx}{dt}+\frac{\partial z}{\partial y}·\frac{dy}{dt}.$

In this diagram, the leftmost corner corresponds to $z=f\left(\mathrm{10},y\right).$ Since $f$ has two independent variables, there are 2 lines coming from this corner. The upper branch corresponds to the variable $\mathrm{10}$ and the lower co-operative corresponds to the variable $y.$ Since each of these variables is and then dependent on one variable $t,$ i co-operative and so comes from $x$ and 1 co-operative comes from $y.$ Concluding, each of the branches on the far right has a label that represents the path traveled to reach that co-operative. The top co-operative is reached by post-obit the $\mathrm{ten}$ branch, and so the $t$ co-operative; therefore, it is labeled $\left(\partial z\text{/}\partial x\right)\times \left(dx\text{/}dt\right).$ The bottom branch is like: start the $y$ branch, then the $t$ co-operative. This branch is labeled $\left(\partial z\text{/}\partial y\right)\times \left(dy\text{/}dt\right).$ To get the formula for $dz\text{/}dt,$ add all the terms that appear on the rightmost side of the diagram. This gives united states of america Equation iv.29.

In Chain Rule for Ii Independent Variables, $z=f\left(x,y\right)$ is a function of $x\text{and}y,$ and both $x=\mathrm{thou}\left(u,v\right)$ and $y=h\left(u,v\right)$ are functions of the contained variables $u\text{and}v.$

Theorem 4.9

Chain Rule for 2 Independent Variables

Suppose $x=\mathrm{one\; thousand}\left(u,5\right)$ and $y=h\left(u,v\right)$ are differentiable functions of $u$ and $v,$ and $z=f\left(x,y\right)$ is a differentiable part of $\mathrm{ten}\text{and}y.$ And then, $z=f\left(g\left(u,v\right),h\left(u,v\right)\right)$ is a differentiable function of $u\text{and}v,$ and

$$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$$

(4.31)

and

$$\frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial \mathrm{five}}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}.$$

(four.32)

We tin describe a tree diagram for each of these formulas as well every bit follows.

Effigy iv.35 Tree diagram for $\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}·\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}·\frac{\partial y}{\partial u}$ and $\frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}·\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}·\frac{\partial y}{\partial \mathrm{five}}.$

To derive the formula for $\partial z\text{/}\partial u,$ outset from the left side of the diagram, then follow merely the branches that cease with $u$ and add together the terms that appear at the finish of those branches. For the formula for $\partial z\text{/}\partial v,$ follow merely the branches that terminate with $v$ and add the terms that appear at the stop of those branches.

There is an important difference between these 2 concatenation dominion theorems. In Chain Rule for One Contained Variable, the left-mitt side of the formula for the derivative is not a partial derivative, but in Chain Rule for Two Independent Variables it is. The reason is that, in Chain Rule for One Independent Variable, $z$ is ultimately a part of $t$ alone, whereas in Concatenation Rule for Two Independent Variables, $z$ is a function of both $u\text{and}\mathrm{five}.$

Instance 4.27

Using the Chain Rule for Two Variables

Summate $\partial z\text{/}\partial u$ and $\partial z\text{/}\partial \mathrm{five}$ using the following functions:

$$z=f\left(\mathrm{10},y\right)=3x^2-\mathrm{two}xy+y^2,x=x\left(u,5\right)=3u+\mathrm{ii}v,y=y\left(u,5\right)=4u-\mathrm{five}.$$

Checkpoint iv.24

Summate $\partial z\text{/}\partial u$ and $\partial z\text{/}\partial v$ given the following functions:

$$z=f\left(x,y\right)=\frac{2\mathrm{ten}-y}{x+3y},x\left(u,\mathrm{five}\right)=e^{2u}\text{cos}\mathrm{iii}v,y\left(u,v\right)={\mathrm{due\; east}}^{2u}\text{sin}3v.$$

The Generalized Chain Rule

Now that nosotros've run into how to extend the original chain rule to functions of two variables, it is natural to ask: Tin can nosotros extend the dominion to more than two variables? The respond is yeah, as the generalized chain rule states.

Theorem iv.10

Generalized Chain Dominion

Permit $w=f\left({\mathrm{10}}_1,x_2\text{,…,}{x}_{\mathrm{one\; thousand}}\right)$ exist a differentiable function of $g$ independent variables, and for each $i\in \left\{\mathrm{one}\text{,…,}m\right\},$ let $x_i={\mathrm{10}}_i(t_{\mathrm{i}},t_2\text{,…,}{t}_{\mathrm{northward}})$ be a differentiable office of $n$ independent variables. And then

$$\frac{\partial \mathrm{due\; west}}{\partial t_j}=\frac{\partial \mathrm{due\; west}}{\partial {\mathrm{ten}}_1}\frac{\partial {\mathrm{10}}_1}{\partial t_j}+\frac{\partial w}{\partial x_2}\frac{\partial x_2}{\partial t_j}+\text{⋯}+\frac{\partial \mathrm{westward}}{\partial {\mathrm{ten}}_{\mathrm{grand}}}\frac{\partial {\mathrm{10}}_{\mathrm{chiliad}}}{\partial t_j}$$

(iv.33)

for any $j\in \{1,2\text{,…,}n\}.$

In the next example we calculate the derivative of a function of three independent variables in which each of the three variables is dependent on two other variables.

Example 4.28

Using the Generalized Chain Dominion

Calculate $\partial w\text{/}\partial u$ and $\partial w\text{/}\partial v$ using the following functions:

$$\begin{array}{}\\ \hfill \mathrm{due\; west}& =\hfill & f\left(x,y,z\right)=\mathrm{three}{x}^2-2\mathrm{10}y+4z^2\hfill \\ \hfill x& =\hfill & x\left(u,v\right)=e^u\text{sin}v\hfill \\ \hfill y& =\hfill & y\left(u,v\right)={\mathrm{east}}^u\text{cos}v\hfill \\ \hfill z& =\hfill & z\left(u,v\right)=e^u.\hfill \end{array}$$

Checkpoint 4.25

Calculate $\partial \mathrm{west}\text{/}\partial u$ and $\partial w\text{/}\partial v$ given the post-obit functions:

$$\begin{array}{}\\ \\ \hfill w& =\hfill & f\left(\mathrm{10},y,z\right)=\frac{x+2y-4z}{\mathrm{ii}\mathrm{10}-y+3z}\hfill \\ \hfill \mathrm{10}& =\hfill & \mathrm{10}\left(u,v\right)={\mathrm{eastward}}^{2u}\text{cos}3v\hfill \\ \hfill y& =\hfill & y\left(u,v\right)={\mathrm{due\; east}}^{\mathrm{two}u}\text{sin}3v\hfill \\ \hfill z& =\hfill & z\left(u,\mathrm{five}\right)=e^{2u}.\hfill \end{array}$$

Example 4.29

Cartoon a Tree Diagram

Create a tree diagram for the example when

$$w=f\left(x,y,z\right),\mathrm{ten}=x\left(t,u,v\right),y=y\left(t,u,v\right),z=z\left(t,u,\mathrm{five}\right)$$

and write out the formulas for the three partial derivatives of $\mathrm{due\; west}.$

Checkpoint four.26

Create a tree diagram for the case when

$$w=f\left(\mathrm{10},y\right),x=x\left(t,u,5\right),y=y\left(t,u,v\right)$$

and write out the formulas for the three partial derivatives of $w.$

Implicit Differentiation

Recall from Implicit Differentiation that implicit differentiation provides a method for finding $dy\text{/}d\mathrm{10}$ when $y$ is defined implicitly every bit a function of $x.$ The method involves differentiating both sides of the equation defining the role with respect to $\mathrm{10},$ then solving for $dy\text{/}d\mathrm{10}.$ Partial derivatives provide an alternative to this method.

Consider the ellipse divers by the equation ${\mathrm{ten}}^{\mathrm{two}}+\mathrm{three}{y}^2+4y-\mathrm{iv}=0$ every bit follows.

Figure 4.37 Graph of the ellipse divers by $x^{\mathrm{ii}}+3y^{\mathrm{ii}}+4y-4=0.$

This equation implicitly defines $y$ as a office of $x.$ As such, we can discover the derivative $dy\text{/}dx$ using the method of implicit differentiation:

$$\begin{array}{ccc}\hfill \frac{d}{d\mathrm{ten}}\left({\mathrm{10}}^{\mathrm{two}}+3y^{\mathrm{two}}+\mathrm{four}y-\mathrm{iv}\right)& =\hfill & \frac{d}{dx}\left(0\right)\hfill \\ \hfill \mathrm{two}x+\mathrm{half\; dozen}y\frac{dy}{dx}+\mathrm{four}\frac{dy}{dx}& =\hfill & 0\hfill \\ \hfill \left(6y+4\right)\frac{dy}{dx}& =\hfill & \mathrm{-2}x\hfill \\ \hfill \frac{dy}{dx}& =\hfill & -\frac{\mathrm{10}}{\mathrm{three}y+2}.\hfill \end{array}$$

Nosotros can too define a function $z=f\left(x,y\right)$ by using the left-hand side of the equation defining the ellipse. Then $f\left(x,y\right)=x^{\mathrm{two}}+3y^{\mathrm{two}}+4y-4.$ The ellipse $x^2+3y^{\mathrm{two}}+4y-\mathrm{iv}=0$ tin can then be described by the equation $f\left(\mathrm{10},y\right)=0.$ Using this office and the post-obit theorem gives united states an alternative approach to computing $dy\text{/}dx.$

Theorem 4.eleven

Implicit Differentiation of a Office of 2 or More Variables

Suppose the function $z=f\left(\mathrm{10},y\right)$ defines $y$ implicitly as a function $y=\mathrm{yard}\left(\mathrm{10}\right)$ of $x$ via the equation $f\left(x,y\right)=0.$ So

$$\frac{dy}{dx}=-\frac{\partial f\text{/}\partial x}{\partial f\text{/}\partial y}$$

(4.34)

provided $f_y\left(\mathrm{10},y\right)\ne 0.$

If the equation $f\left(x,y,z\right)=0$ defines $z$ implicitly equally a differentiable function of $\mathrm{ten}\text{and}y,$ and so

$$\frac{\partial z}{\partial x}=-\frac{\partial f\text{/}\partial x}{\partial f\text{/}\partial z}\text{and}\frac{\partial z}{\partial y}=-\frac{\partial f\text{/}\partial y}{\partial f\text{/}\partial z}$$

(iv.35)

as long as $f_z\left(x,y,z\right)\ne 0.$

Equation 4.34 is a direct consequence of Equation 4.31. In particular, if we assume that $y$ is defined implicitly as a function of $x$ via the equation $f\left(\mathrm{10},y\right)=0,$ we tin use the chain dominion to find $dy\text{/}dx\text{:}$

$$\begin{array}{ccc}\hfill \frac{d}{dx}f\left(x,y\right)& =\hfill & \frac{d}{dx}\left(0\right)\hfill \\ \hfill \frac{\partial f}{\partial x}·\frac{dx}{dx}+\frac{\partial f}{\partial y}·\frac{dy}{d\mathrm{10}}& =\hfill & 0\hfill \\ \hfill \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}·\frac{dy}{d\mathrm{10}}& =\hfill & 0.\hfill \end{array}$$

Solving this equation for $dy\text{/}d\mathrm{10}$ gives Equation 4.34. Equation iv.35 tin can be derived in a similar fashion.

Allow'southward now return to the problem that we started before the previous theorem. Using Implicit Differentiation of a Office of Two or More Variables and the function $f\left(\mathrm{10},y\right)={\mathrm{10}}^2+\mathrm{three}{y}^{\mathrm{two}}+4y-4,$ we obtain

$$\begin{array}{c}\frac{\partial f}{\partial x}=2x\hfill \\ \frac{\partial f}{\partial y}=6y+4.\hfill \end{array}$$

Then Equation 4.34 gives

$$\frac{dy}{dx}=-\frac{\partial f\text{/}\partial \mathrm{ten}}{\partial f\text{/}\partial y}=-\frac{2\mathrm{ten}}{6y+4}=-\frac{\mathrm{ten}}{3y+2},$$

which is the same result obtained past the earlier use of implicit differentiation.

Example iv.30

Implicit Differentiation past Partial Derivatives

1. Calculate $dy\text{/}d\mathrm{10}$ if $y$ is defined implicitly equally a function of $x$ via the equation $3x^{\mathrm{two}}-\mathrm{ii}xy+y^2+4x-6y-11=0.$ What is the equation of the tangent line to the graph of this curve at point $\left(\mathrm{ii},1\right)?$
2. Calculate $\partial z\text{/}\partial \mathrm{10}$ and $\partial z\text{/}\partial y,$ given $x^2{e}^y-yz{e}^x=0.$

Checkpoint 4.27

Find $dy\text{/}dx$ if $y$ is defined implicitly as a function of $x$ by the equation $x^{\mathrm{two}}+\mathrm{ten}y-y^2+7\mathrm{ten}-3y-26=0.$ What is the equation of the tangent line to the graph of this bend at point $\left(\mathrm{three},\mathrm{-2}\right)?$

Section four.five Exercises

For the following exercises, use the information provided to solve the problem.

215.

Let $\mathrm{westward}\left(\mathrm{10},y,z\right)=xy\text{cos}z,$ where $\mathrm{ten}=t,y=t^2,$ and $z=\text{arcsin}t.$ Notice $\frac{d\mathrm{west}}{dt}.$

216 .

Let $\mathrm{westward}\left(t,5\right)=e^{t\mathrm{five}}$ where $t=r+\mathrm{due\; south}$ and $5=rs.$ Find $\frac{\partial w}{\partial r}$ and $\frac{\partial \mathrm{westward}}{\partial \mathrm{south}}.$

217.

If $w=5{\mathrm{10}}^2+2y^2,\mathrm{10}=\mathrm{-3}s+t,$ and $y=\mathrm{south}-4t,$ discover $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t}.$

218 .

If $w=x{y}^2,x=\mathrm{v}\text{cos}(2t),$ and $y=5\text{sin}(2t),$ find $\frac{dw}{dt}.$

219.

If $f(x,y)=xy,x=r\text{cos}\theta ,$ and $y=r\text{sin}\theta ,$ detect $\frac{\partial f}{\partial r}$ and limited the respond in terms of $r$ and $\theta .$

220 .

Suppose $f(x,y)=\mathrm{10}+y,$ where $x=r\text{cos}\theta$ and $y=r\text{sin}\theta .$ Find $\frac{\partial f}{\partial \theta }.$

For the post-obit exercises, detect $\frac{df}{dt}$ using the chain rule and straight substitution.

221.

$f\left(x,y\right)=x^2+y^{\mathrm{ii}},$ $x=t,y=t^2$

222 .

$f\left(x,y\right)=\sqrt{x^{\mathrm{two}}+y^{\mathrm{two}}},y=t^2,x=t$

223.

$f\left(x,y\right)=xy,x=\mathrm{one}-\sqrt{t},y=1+\sqrt{t}$

224 .

$f\left(x,y\right)=\frac{\mathrm{10}}{y},\mathrm{ten}={\mathrm{east}}^t,y=2e^t$

225.

$f\left(x,y\right)=\text{ln}\left(x+y\right),$ $x={\mathrm{eastward}}^t,y=e^t$

226 .

$f\left(\mathrm{10},y\right)=x^4,$ $x=t,y=t$

227.

Let $w\left(x,y,z\right)=x^{\mathrm{two}}+y^{\mathrm{two}}+z^{\mathrm{two}},$ $x=\text{cos}t,y=\text{sin}t,$ and $z=e^t.$ Limited $w$ as a part of $t$ and detect $\frac{dw}{dt}$ straight. Then, find $\frac{dw}{dt}$ using the chain rule.

228 .

Allow $z=x^{2}y,$ where $x=t^2$ and $y=t^3.$ Find $\frac{dz}{dt}.$

229.

Let $u={\mathrm{due\; east}}^x\text{sin}y,$ where $\mathrm{10}=-ln2t$ and $y=\pi t.$ Discover $\frac{du}{dt}$ when $x=\text{ln}2$ and $y=\frac{\pi }{4}.$

For the following exercises, discover $\frac{dy}{d\mathrm{10}}$ using partial derivatives.

230 .

$\text{sin}\left(6x\right)+\text{tan}\left(\mathrm{eight}y\right)+5=0$

231.

$x^{\mathrm{three}}+y^2\mathrm{10}-\mathrm{iii}=0$

232 .

$\text{sin}\left(\mathrm{ten}+y\right)+\text{cos}\left(\mathrm{10}-y\right)=4$

233.

$x^2-\mathrm{two}xy+y^{\mathrm{four}}=4$

234 .

$\mathrm{ten}{e}^y+y{e}^x-2x^{2}y=0$

235.

$x^{2\text{/}\mathrm{three}}+y^{2\text{/}3}=a^{2\text{/}\mathrm{iii}}$

236 .

$x\text{cos}(xy)+y\text{cos}\mathrm{ten}=2$

237.

${\mathrm{east}}^{xy}+y{e}^y=\mathrm{i}$

238 .

${\mathrm{ten}}^2{y}^{\mathrm{iii}}+\text{cos}y=0$

239.

Find $\frac{dz}{dt}$ using the chain dominion where $z=3{\mathrm{ten}}^{\mathrm{ii}}{y}^{\mathrm{three}},x=t^{\mathrm{iv}},$ and $y=t^{\mathrm{ii}}.$

240 .

Let $z=\mathrm{iii}\text{cos}x-\text{sin}(\mathrm{10}y),x=\frac{1}{t},$ and $y=\mathrm{iii}t.$ Observe $\frac{dz}{dt}.$

241.

Let $z=e^{\mathrm{one}-xy},\mathrm{ten}=t^{\mathrm{i}\text{/}\mathrm{iii}},$ and $y=t^{\mathrm{three}}.$ Find $\frac{dz}{dt}.$

242 .

Find $\frac{dz}{dt}$ by the chain rule where $z={\text{cosh}}^2(xy),x=\frac{1}{\mathrm{ii}}t,$ and $y=e^t.$

243.

Let $z=\frac{x}{y},x=2\text{cos}u,$ and $y=\mathrm{three}\text{sin}5.$ Find $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial 5}.$

244 .

Allow $z=e^{x^{2}y},$ where $x=\sqrt{uv}$ and $y=\frac{1}{v}.$ Detect $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v}.$

245.

If $z=xy{e}^{x\text{/}y},$ $\mathrm{10}=r\text{cos}\theta ,$ and $y=r\text{sin}\theta ,$ find $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial \theta }$ when $r=2$ and $\theta =\frac{\pi }{\mathrm{half-dozen}}.$

246 .

Find $\frac{\partial w}{\partial s}$ if $w=\mathrm{iv}\mathrm{10}+y^{\mathrm{ii}}+z^3,x={\mathrm{eastward}}^{r{s}^2},y=\text{ln}\left(\frac{r+\mathrm{southward}}{t}\right),$ and $z=rs{t}^2.$

247.

If $w=\text{sin}(xyz),x=1-3t,y=e^{1-t},$ and $z=4t,$ find $\frac{\partial \mathrm{due\; west}}{\partial t}.$

For the following exercises, apply this information: A office $f(x,y)$ is said to be homogeneous of degree $n$ if $f(t\mathrm{10},ty)=t^{n}f(\mathrm{10},y).$ For all homogeneous functions of degree $n,$ the following equation is true: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y).$ Prove that the given function is homogeneous and verify that $x\frac{\partial f}{\partial \mathrm{10}}+y\frac{\partial f}{\partial y}=nf(\mathrm{ten},y).$

248 .

$f(x,y)=3{\mathrm{ten}}^{\mathrm{two}}+y^2$

249.

$f(\mathrm{ten},y)=\sqrt{{\mathrm{10}}^{\mathrm{two}}+y^2}$

250 .

$f(x,y)=x^{2}y-\mathrm{two}{y}^3$

251.

The book of a right circular cylinder is given past $V(x,y)=\pi x^{2}y,$ where $\mathrm{10}$ is the radius of the cylinder and y is the cylinder height. Suppose $\mathrm{10}$ and $y$ are functions of $t$ given by $x=\frac{1}{2}t$ and $y=\frac{1}{3}t$ so that $\mathrm{10}\text{and}y$ are both increasing with time. How fast is the volume increasing when $\mathrm{ten}=2$ and $y=\frac{\mathrm{four}}{3}?$

252 .

The pressure $P$ of a gas is related to the volume and temperature by the formula $PV=kT,$ where temperature is expressed in kelvins. Express the force per unit area of the gas as a function of both $\mathrm{Five}$ and $T.$ Find $\frac{dP}{dt}$ when $k=1,$ $\frac{dV}{dt}=2$ cm³/min, $\frac{dT}{dt}=\frac{1}{2}$ 1000/min, $V=\mathrm{twenty}$ cm³, and $T=20\text{°}\text{F}.$

253.

The radius of a right circular cone is increasing at $3$ cm/min whereas the meridian of the cone is decreasing at $2$ cm/min. Find the rate of alter of the volume of the cone when the radius is $\mathrm{xiii}$ cm and the height is $18$ cm.

254 .

The volume of a frustum of a cone is given by the formula $V=\frac{1}{3}\pi z\left(x^2+y^{\mathrm{ii}}+xy\right),$ where $x$ is the radius of the smaller circle, $y$ is the radius of the larger circumvolve, and $z$ is the height of the frustum (meet effigy). Find the charge per unit of alter of the volume of this frustum when $\mathrm{ten}=10\text{in}\text{.,}y=12\text{in.,}\text{and}z=18\text{in}.$ if $\frac{dz}{dt}=-5,\frac{dx}{dt}=\mathrm{i},\frac{dy}{dt}=\mathrm{i}$ (all in/min).

255.

A closed box is in the shape of a rectangular solid with dimensions $x,y,\text{and}z.$ (Dimensions are in inches.) Suppose each dimension is changing at the charge per unit of $\mathrm{0.v}$ in./min. Notice the charge per unit of change of the full surface expanse of the box when $x=2\text{in}\text{.,}y=\mathrm{iii}\text{in.,}\text{and}z=1\text{in}.$

256 .

The total resistance in a circuit that has three individual resistances represented past $x,y,$ and $z$ is given by the formula $R(x,y,z)=\frac{\mathrm{ten}yz}{yz+\mathrm{ten}z+\mathrm{ten}y}.$ Suppose at a given time the $x$ resistance is $100\text{Ω},$ the y resistance is $200\text{Ω},$ and the $z$ resistance is $300\text{Ω}.$ Also, suppose the $\mathrm{10}$ resistance is irresolute at a rate of $2\text{Ω}\text{/}\text{min},$ the $y$ resistance is changing at the rate of $\mathrm{one}\text{Ω}\text{/}\text{min},$ and the $z$ resistance has no alter. Find the rate of change of the full resistance in this excursion at this time.

257.

The temperature $T$ at a point $(x,y)$ is $T(x,y)$ and is measured using the Celsius scale. A fly crawls so that its position after $t$ seconds is given by $x=\sqrt{1+t}$ and $y=\mathrm{two}+\frac{1}{3}t,$ where $x\text{and}y$ are measured in centimeters. The temperature function satisfies $T_x\left(\mathrm{two},\mathrm{three}\right)=4$ and $T_y\left(\mathrm{two},3\right)=3.$ How fast is the temperature increasing on the fly'south path after $3$ sec?

258 .

The $x\text{and}y$ components of a fluid moving in two dimensions are given by the following functions: $u(x,y)=2y$ and $v(\mathrm{10},y)=\mathrm{-2}\mathrm{10};$ $\mathrm{ten}\ge 0;y\ge 0.$ The speed of the fluid at the signal $(x,y)$ is $\mathrm{south}(\mathrm{ten},y)=\sqrt{u{(\mathrm{10},y)}^2+\mathrm{five}{(x,y)}^2}.$ Notice $\frac{\partial \mathrm{south}}{\partial x}$ and $\frac{\partial s}{\partial y}$ using the chain dominion.

259.

Let $u=u\left(x,y,z\right),$ where $\mathrm{10}=\mathrm{10}(\mathrm{westward},t),y=y(\mathrm{west},t),z=z(w,t),\mathrm{west}=w(r,\mathrm{southward}),\text{and}t=t(r,\mathrm{south}).$ Use a tree diagram and the chain rule to find an expression for $\frac{\partial u}{\partial r}.$

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Source: https://openstax.org/books/calculus-volume-3/pages/4-5-the-chain-rule

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