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The Chain Rule

State the chain rules for one or two independent variables.

Chain Rules for One or Two Independent Variables

Recall that the chain rule for the derivative of a composite of two functions can be written in the form

\[\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x).\]

In this equation, both \(f(x)\) and \(g(x)\) are functions of one variable. Now suppose that \(f\) is a function of two variables and \(g\) is a function of one variable. Or perhaps they are both functions of two variables, or even more. How would we calculate the derivative in these cases? The following theorem gives us the answer for the case of one independent variable.

Condensed: the full section is in OpenStax Calculus Volume 3.

The Generalized Chain Rule

Now that we’ve see how to extend the original chain rule to functions of two variables, it is natural to ask: Can we extend the rule to more than two variables? The answer is yes, as the generalized chain rule states.

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

Try it.

Create a tree diagram for the case when

\[w=f(x,y,z),x=x(t,u,v),y=y(t,u,v),z=z(t,u,v)\]

and write out the formulas for the three partial derivatives of \(w.\)

Solution

Starting from the left, the function \(f\) has three independent variables: \(x,y,\text{and}\ z.\) Therefore, three branches must be emanating from the first node. Each of these three branches also has three branches, for each of the variables \(t,u,\ \text{and}\ v.\)

The three formulas are

\[\begin{array}{l}\frac{∂w}{∂t}=\frac{∂w}{∂x}\ \frac{∂x}{∂t}+\frac{∂w}{∂y}\ \frac{∂y}{∂t}+\frac{∂w}{∂z}\ \frac{∂z}{∂t} \\ \frac{∂w}{∂u}=\frac{∂w}{∂x}\ \frac{∂x}{∂u}+\frac{∂w}{∂y}\ \frac{∂y}{∂u}+\frac{∂w}{∂z}\ \frac{∂z}{∂u} \\ \frac{∂w}{∂v}=\frac{∂w}{∂x}\ \frac{∂x}{∂v}+\frac{∂w}{∂y}\ \frac{∂y}{∂v}+\frac{∂w}{∂z}\ \frac{∂z}{∂v}.\end{array}\]

Condensed: the full section is in OpenStax Calculus Volume 3.

Implicit Differentiation

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

Consider the ellipse defined by the equation \({x}^{2}+3{y}^{2}+4y-4=0\) as follows.

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

\[\begin{array}{lll}\frac{d}{dx}({x}^{2}+3{y}^{2}+4y-4) & = & \frac{d}{dx}(0) \\ 2x+6y\frac{dy}{dx}+4\frac{dy}{dx} & = & 0 \\ (6y+4)\frac{dy}{dx} & = & -2x \\ \frac{dy}{dx} & = & -\frac{x}{3y+2}.\end{array}\]

We can also define a function \(z=f(x,y)\) by using the left-hand side of the equation defining the ellipse. Then \(f(x,y)={x}^{2}+3{y}^{2}+4y-4.\) The ellipse \({x}^{2}+3{y}^{2}+4y-4=0\) can then be described by the equation \(f(x,y)=0.\) Using this function and the following theorem gives us an alternative approach to calculating \(dy\text{/}dx.\)

is a direct consequence of . In particular, if we assume that \(y\) is defined implicitly as a function of \(x\) via the equation \(f(x,y)=0,\) we can apply the chain rule to find \(dy\text{/}dx\text{:}\)

\[\begin{array}{lll}\frac{d}{dx}f(x,y) & = & \frac{d}{dx}(0) \\ \frac{∂f}{∂x}\cdot \frac{dx}{dx}+\frac{∂f}{∂y}\cdot \frac{dy}{dx} & = & 0 \\ \frac{∂f}{∂x}+\frac{∂f}{∂y}\cdot \frac{dy}{dx} & = & 0.\end{array}\]

Solving this equation for \(dy\text{/}dx\) gives . can be derived in a similar fashion.

Let’s now return to the problem that we started before the previous theorem. Using and the function \(f(x,y)={x}^{2}+3{y}^{2}+4y-4,\) we obtain

\[\begin{array}{l}\frac{∂f}{∂x}=2x \\ \frac{∂f}{∂y}=6y+4.\end{array}\]\[\frac{dy}{dx}=-\frac{∂f\text{/}∂x}{∂f\text{/}∂y}=-\frac{2x}{6y+4}=-\frac{x}{3y+2},\]

Condensed: the full section is in OpenStax Calculus Volume 3.

Key Concepts

  • The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables.
  • Tree diagrams are useful for deriving formulas for the chain rule for functions of more than one variable, where each independent variable also depends on other variables.

Key Equations

Chain rule, one independent variable\(\frac{dz}{dt}=\frac{∂z}{∂x}\cdot \frac{dx}{dt}+\frac{∂z}{∂y}\cdot \frac{dy}{dt}\)
Chain rule, two independent variables\(\frac{dz}{du}=\frac{∂z}{∂x}\frac{∂x}{∂u}+\frac{∂z}{∂y}\frac{∂y}{∂u}\)
\(\frac{dz}{dv}=\frac{∂z}{∂x}\frac{∂x}{∂v}+\frac{∂z}{∂y}\frac{∂y}{∂v}\)
Generalized chain rule\(\frac{∂w}{∂{t}_{j}}=\frac{∂w}{∂{x}_{1}}\ \frac{∂{x}_{1}}{∂{t}_{j}}+\frac{∂w}{∂{x}_{2}}\ \frac{∂{x}_{1}}{∂{t}_{j}}+\cdots +\frac{∂w}{∂{x}_{m}}\ \frac{∂{x}_{m}}{∂{t}_{j}}\)

The Chain Rule

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

For the following exercises, find \(\frac{df}{dt}\) using the chain rule and direct substitution.

For the following exercises, find \(\frac{dy}{dx}\) using partial derivatives.

For the following exercises, use this information: A function \(f(x,y)\) is said to be homogeneous of degree \(n\) if \(f(tx,ty)={t}^{n}f(x,y).\) For all homogeneous functions of degree \(n,\) the following equation is true: \(x\frac{∂f}{∂x}+y\frac{∂f}{∂y}=nf(x,y).\) Show that the given function is homogeneous and verify that \(x\frac{∂f}{∂x}+y\frac{∂f}{∂y}=nf(x,y).\)

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  1. Calculate \(dz\text{/}dt\) for each of the following functions:

    1. \(z=f(x,y)=4{x}^{2}+3{y}^{2},x=x(t)=\text{sin}\ t,y=y(t)=\text{cos}\ t\)
    2. \(z=f(x,y)=\sqrt{{x}^{2}-{y}^{2}},x=x(t)={e}^{2t},y=y(t)={e}^{\text{-}t}\)
    Antwort aufdecken
    1. To use the chain rule, we need four quantities, \(∂z\text{/}∂x,∂z\text{/}∂y,dx\text{/}dt,\) and \(dy\text{/}dt\text{:}\)
      \[\begin{array}{llll}\frac{∂z}{∂x}=8x & & & \frac{∂z}{∂y}=6y \\ \frac{dx}{dt}=\text{cos}\ t & & & \frac{dy}{dt}=\text{-}\text{sin}\ t\end{array}\]
      Now, we substitute each of these into :
      \[\begin{array}{ll}\frac{dz}{dt} & =\frac{∂z}{∂x}\cdot \frac{dx}{dt}+\frac{∂z}{∂y}\cdot \frac{dy}{dt} \\ & =(8x)(\text{cos}\ t)+(6y)(\text{-}\text{sin}\ t) \\ & =8x\ \text{cos}\ t-6y\ \text{sin}\ t.\end{array}\]
      This answer has three variables in it. To reduce it to one variable, use the fact that \(x(t)=\text{sin}\ t\ \text{and}\ y(t)=\text{cos}\ t.\) We obtain
      \[\begin{array}{ll}\frac{dz}{dt} & =8x\ \text{cos}\ t-6y\ \text{sin}\ t \\ & =8(\text{sin}\ t)\text{cos}\ t-6(\text{cos}\ t)\text{sin}\ t \\ & =2\ \text{sin}\ t\ \text{cos}\ t.\end{array}\]
      This derivative can also be calculated by first substituting \(x(t)\) and \(y(t)\) into \(f(x,y),\) then differentiating with respect to \(t\text{:}\)
      \[\begin{array}{ll}z & =f(x,y) \\ & =f(x(t),y(t)) \\ & =4{(x(t))}^{2}+3{(y(t))}^{2} \\ & =4{\text{sin}}^{2}t+3{\text{cos}}^{2}t.\end{array}\]
      Then
      \[\begin{array}{ll}\frac{dz}{dt} & =2(4\ \text{sin}\ t)(\text{cos}\ t)+2(3\ \text{cos}\ t)(\text{-}\text{sin}\ t) \\ & =8\ \text{sin}\ t\ \text{cos}\ t-6\ \text{sin}\ t\ \text{cos}\ t \\ & =2\ \text{sin}\ t\ \text{cos}\ t,\end{array}\]
      which is the same solution. However, it may not always be this easy to differentiate in this form.
    2. To use the chain rule, we again need four quantities, \(∂z\text{/}∂x,∂z\text{/}dy,dx\text{/}dt,\) and \(dy\text{/}dt\text{:}\)
      \[\begin{array}{llll}\frac{∂z}{∂x}=\frac{x}{\sqrt{{x}^{2}-{y}^{2}}} & & & \frac{∂z}{∂y}=\frac{\text{-}y}{\sqrt{{x}^{2}-{y}^{2}}} \\ \frac{dx}{dt}=2{e}^{2t} & & & \frac{dy}{dt}=\text{-}{e}^{\text{-}t}.\end{array}\]
      We substitute each of these into :
      \[\begin{array}{ll}\frac{dz}{dt} & =\frac{∂z}{∂x}\cdot \frac{dx}{dt}+\frac{∂z}{∂y}\cdot \frac{dy}{dt} \\ & =(\frac{x}{\sqrt{{x}^{2}-{y}^{2}}})(2{e}^{2t})+(\frac{\text{-}y}{\sqrt{{x}^{2}-{y}^{2}}})(\text{-}{e}^{\text{-}t}) \\ & =\frac{2x{e}^{2t}+y{e}^{\text{-}t}}{\sqrt{{x}^{2}-{y}^{2}}}.\end{array}\]
      To reduce this to one variable, we use the fact that \(x(t)={e}^{2t}\) and \(y(t)={e}^{\text{-}t}.\) Therefore,
      \[\begin{array}{ll}\frac{dz}{dt} & =\frac{2x{e}^{2t}+y{e}^{\text{-}t}}{\sqrt{{x}^{2}-{y}^{2}}} \\ & =\frac{2({e}^{2t}){e}^{2t}+({e}^{\text{-}t}){e}^{\text{-}t}}{\sqrt{{e}^{4t}-{e}^{-2t}}} \\ & =\frac{2{e}^{4t}+{e}^{-2t}}{\sqrt{{e}^{4t}-{e}^{-2t}}}.\end{array}\]
      To eliminate negative exponents, we multiply the top by \({e}^{2t}\) and the bottom by \(\sqrt{{e}^{4t}}\text{:}\)
      \[\begin{array}{ll}\frac{dz}{dt} & =\frac{2{e}^{4t}+{e}^{-2t}}{\sqrt{{e}^{4t}-{e}^{-2t}}}\cdot \frac{{e}^{2t}}{\sqrt{{e}^{4t}}} \\ & =\frac{2{e}^{6t}+1}{\sqrt{{e}^{8t}-{e}^{2t}}} \\ & =\frac{2{e}^{6t}+1}{\sqrt{{e}^{2t}({e}^{6t}-1)}} \\ & =\frac{2{e}^{6t}+1}{{e}^{t}\sqrt{{e}^{6t}-1}}.\end{array}\]
      Again, this derivative can also be calculated by first substituting \(x(t)\) and \(y(t)\) into \(f(x,y),\) then differentiating with respect to \(t\text{:}\)
      \[\begin{array}{ll}z & =f(x,y) \\ & =f(x(t),y(t)) \\ & =\sqrt{{(x(t))}^{2}-{(y(t))}^{2}} \\ & =\sqrt{{e}^{4t}-{e}^{-2t}} \\ & ={({e}^{4t}-{e}^{-2t})}^{1\text{/}2}.\end{array}\]
      Then
      \[\begin{array}{ll}\frac{dz}{dt} & =\frac{1}{2}{({e}^{4t}-{e}^{-2t})}^{\text{-}1\text{/}2}(4{e}^{4t}+2{e}^{-2t}) \\ & =\frac{2{e}^{4t}+{e}^{-2t}}{\sqrt{{e}^{4t}-{e}^{-2t}}}.\end{array}\]
      This is the same solution.

  2. Calculate \(dz\text{/}dt\) given the following functions. Express the final answer in terms of \(t.\)

    \[z=f(x,y)={x}^{2}-3xy+2{y}^{2},x=x(t)=3\ \text{sin}\ 2t,y=y(t)=4\ \text{cos}\ 2t\]
    Antwort aufdecken

    \(\begin{array}{ll}\frac{dz}{dt} & =\frac{∂f}{∂x}\ \frac{dx}{dt}+\frac{∂f}{∂y}\ \frac{dy}{dt} \\ & =(2x-3y)(6\ \text{cos}\ 2t)+(-3x+4y)(-8\ \text{sin}\ 2t) \\ & =-92\ \text{sin}\ 2t\ \text{cos}\ 2t-72({\text{cos}}^{2}2t-{\text{sin}}^{2}2t) \\ & =-46\ \text{sin}\ 4t-72\ \text{cos}\ 4t.\end{array}\)

  3. Calculate \(∂z\text{/}∂u\) and \(∂z\text{/}∂v\) using the following functions:

    \[z=f(x,y)=3{x}^{2}-2xy+{y}^{2},x=x(u,v)=3u+2v,y=y(u,v)=4u-v.\]
    Antwort aufdecken

    To implement the chain rule for two variables, we need six partial derivatives, \(∂z\text{/}∂x,∂z\text{/}∂y,∂x\text{/}∂u,∂x\text{/}∂v,∂y\text{/}∂u,\) and \(∂y\text{/}∂v\text{:}\)

    \[\begin{array}{llll}\frac{∂z}{∂x}=6x-2y & & & \frac{∂z}{∂y}=-2x+2y \\ \frac{∂x}{∂u}=3 & & & \frac{∂x}{∂v}=2 \\ \frac{∂y}{∂u}=4 & & & \frac{∂y}{∂v}=-1.\end{array}\]

    To find \(∂z\text{/}∂u,\) we use :

    \[\begin{array}{ll}\frac{∂z}{∂u} & =\frac{∂z}{∂x}\frac{∂x}{∂u}+\frac{∂z}{∂y}\frac{∂y}{∂u} \\ & =3(6x-2y)+4(-2x+2y) \\ & =10x+2y.\end{array}\]

    Next, we substitute \(x(u,v)=3u+2v\) and \(y(u,v)=4u-v\text{:}\)

    \[\begin{array}{ll}\frac{∂z}{∂u} & =10x+2y \\ & =10(3u+2v)+2(4u-v) \\ & =38u+18v.\end{array}\]

    To find \(∂z\text{/}∂v,\) we use :

    \[\begin{array}{ll}\frac{∂z}{∂v} & =\frac{∂z}{∂x}\ \frac{∂x}{∂v}+\frac{∂z}{∂y}\ \frac{∂y}{∂v} \\ & =2(6x-2y)+(-1)(-2x+2y) \\ & =14x-6y.\end{array}\]

    Then we substitute \(x(u,v)=3u+2v\) and \(y(u,v)=4u-v\text{:}\)

    \[\begin{array}{ll}\frac{∂z}{∂v} & =14x-6y \\ & =14(3u+2v)-6(4u-v) \\ & =18u+34v.\end{array}\]

  4. Calculate \(∂z\text{/}∂u\) and \(∂z\text{/}∂v\) given the following functions:

    \[z=f(x,y)=\frac{2x-y}{x+3y},x(u,v)={e}^{2u}\text{cos}\ 3v,y(u,v)={e}^{2u}\text{sin}\ 3v.\]
    Antwort aufdecken

    \(\frac{∂z}{∂u}=0,\ \frac{∂z}{∂v}=\frac{-21}{{(3\ \text{sin}\ 3v+\text{cos}\ 3v)}^{2}}\)

  5. Calculate \(∂w\text{/}∂u\) and \(∂w\text{/}∂v\) using the following functions:

    \[\begin{array}{lll} \\ w & = & f(x,y,z)=3{x}^{2}-2xy+4{z}^{2} \\ x & = & x(u,v)={e}^{u}\text{sin}\ v \\ y & = & y(u,v)={e}^{u}\text{cos}\ v \\ z & = & z(u,v)={e}^{u}.\end{array}\]
    Antwort aufdecken

    The formulas for \(∂w\text{/}∂u\) and \(∂w\text{/}∂v\) are

    \[\begin{array}{l}\frac{∂w}{∂u}=\frac{∂w}{∂x}\cdot \frac{∂x}{∂u}+\frac{∂w}{∂y}\cdot \frac{∂y}{∂u}+\frac{∂w}{∂z}\cdot \frac{∂z}{∂u} \\ \frac{∂w}{∂v}=\frac{∂w}{∂x}\cdot \frac{∂x}{∂v}+\frac{∂w}{∂y}\cdot \frac{∂y}{∂v}+\frac{∂w}{∂z}\cdot \frac{∂z}{∂v}.\end{array}\]

    Therefore, there are nine different partial derivatives that need to be calculated and substituted. We need to calculate each of them:

    \[\begin{array}{lllllllllllll}\frac{∂w}{∂x} & = & 6x-2y & & & \frac{∂w}{∂y} & = & -2x & & & \frac{∂w}{∂z} & = & 8z \\ \frac{∂x}{∂u} & = & {e}^{u}\text{sin}\ v & & & \frac{∂y}{∂u} & = & {e}^{u}\text{cos}\ v & & & \frac{∂z}{∂u} & = & {e}^{u} \\ \frac{∂x}{∂v} & = & {e}^{u}\text{cos}\ v & & & \frac{∂y}{∂v} & = & \text{-}{e}^{u}\text{sin}\ v & & & \frac{∂z}{∂v} & = & 0.\end{array}\]

    Now, we substitute each of them into the first formula to calculate \(∂w\text{/}∂u\text{:}\)

    \[\begin{array}{ll}\frac{∂w}{∂u} & =\frac{∂w}{∂x}\cdot \frac{∂x}{∂u}+\frac{∂w}{∂y}\cdot \frac{∂y}{∂u}+\frac{∂w}{∂z}\cdot \frac{∂z}{∂u} \\ & =(6x-2y){e}^{u}\text{sin}\ v-2x{e}^{u}\text{cos}\ v+8z{e}^{u},\end{array}\]

    then substitute \(x(u,v)={e}^{u}\text{sin}\ v,y(u,v)={e}^{u}\text{cos}\ v,\) and \(z(u,v)={e}^{u}\) into this equation:

    \[\begin{array}{ll}\frac{∂w}{∂u} & =(6x-2y){e}^{u}\text{sin}\ v-2x{e}^{u}\text{cos}\ v+8z{e}^{u} \\ & =(6{e}^{u}\text{sin}\ v-2{e}^{u}\text{cos}\ v){e}^{u}\text{sin}\ v-2({e}^{u}\text{sin}\ v){e}^{u}\text{cos}\ v+8{e}^{2u} \\ & =6{e}^{2u}{\text{sin}}^{2}v-4{e}^{2u}\text{sin}\ v\ \text{cos}\ v+8{e}^{2u} \\ & =2{e}^{2u}(3\ {\text{sin}}^{2}v-2\ \text{sin}\ v\ \text{cos}\ v+4).\end{array}\]

    Next, we calculate \(∂w\text{/}∂v\text{:}\)

    \[\begin{array}{ll}\frac{∂w}{∂v} & =\frac{∂w}{∂x}\cdot \frac{∂x}{∂v}+\frac{∂w}{∂y}\cdot \frac{∂y}{∂v}+\frac{∂w}{∂z}\cdot \frac{∂z}{∂v} \\ & =(6x-2y){e}^{u}\text{cos}\ v-2x(\text{-}{e}^{u}\text{sin}\ v)+8z(0),\end{array}\]

    then we substitute \(x(u,v)={e}^{u}\text{sin}\ v,y(u,v)={e}^{u}\text{cos}\ v,\) and \(z(u,v)={e}^{u}\) into this equation:

    \[\begin{array}{ll}\frac{∂w}{∂v} & =(6x-2y){e}^{u}\text{cos}\ v-2x(\text{-}{e}^{u}\text{sin}\ v) \\ & =(6{e}^{u}\text{sin}\ v-2{e}^{u}\text{cos}\ v){e}^{u}\text{cos}\ v+2({e}^{u}\text{sin}\ v)({e}^{u}\text{sin}\ v) \\ & =2{e}^{2u}{\text{sin}}^{2}v+6{e}^{2u}\text{sin}\ v\ \text{cos}\ v-2{e}^{2u}{\text{cos}}^{2}v \\ & =2{e}^{2u}(v+3\text{sin}\text{sin }v\text{cos}\text{cos }v-v).\end{array}\]

  6. Calculate \(∂w\text{/}∂u\) and \(∂w\text{/}∂v\) given the following functions:

    \[\begin{array}{lll} \\ \\ w & = & f(x,y,z)=\frac{x+2y-4z}{2x-y+3z} \\ x & = & x(u,v)={e}^{2u}\text{cos}\ 3v \\ y & = & y(u,v)={e}^{2u}\text{sin}\ 3v \\ z & = & z(u,v)={e}^{2u}.\end{array}\]
    Antwort aufdecken

    \(\begin{array}{l}\frac{∂w}{∂u}=0 \\ \frac{∂w}{∂v}=\frac{15-33\ \text{sin}\ 3v+6\ \text{cos}\ 3v}{{(3+2\ \text{cos}\ 3v-\text{sin}\ 3v)}^{2}}\end{array}\)

  7. Create a tree diagram for the case when

    \[w=f(x,y,z),x=x(t,u,v),y=y(t,u,v),z=z(t,u,v)\]

    and write out the formulas for the three partial derivatives of \(w.\)

    Antwort aufdecken

    Starting from the left, the function \(f\) has three independent variables: \(x,y,\text{and}\ z.\) Therefore, three branches must be emanating from the first node. Each of these three branches also has three branches, for each of the variables \(t,u,\ \text{and}\ v.\)

    The three formulas are

    \[\begin{array}{l}\frac{∂w}{∂t}=\frac{∂w}{∂x}\ \frac{∂x}{∂t}+\frac{∂w}{∂y}\ \frac{∂y}{∂t}+\frac{∂w}{∂z}\ \frac{∂z}{∂t} \\ \frac{∂w}{∂u}=\frac{∂w}{∂x}\ \frac{∂x}{∂u}+\frac{∂w}{∂y}\ \frac{∂y}{∂u}+\frac{∂w}{∂z}\ \frac{∂z}{∂u} \\ \frac{∂w}{∂v}=\frac{∂w}{∂x}\ \frac{∂x}{∂v}+\frac{∂w}{∂y}\ \frac{∂y}{∂v}+\frac{∂w}{∂z}\ \frac{∂z}{∂v}.\end{array}\]

  8. Create a tree diagram for the case when

    \[w=f(x,y),x=x(t,u,v),y=y(t,u,v)\]

    and write out the formulas for the three partial derivatives of \(w.\)

    Antwort aufdecken

    \(\begin{array}{l}\frac{∂w}{∂t}=\frac{∂w}{∂x}\ \frac{∂x}{∂t}+\frac{∂w}{∂y}\ \frac{∂y}{∂t} \\ \frac{∂w}{∂u}=\frac{∂w}{∂x}\ \frac{∂x}{∂u}+\frac{∂w}{∂y}\ \frac{∂y}{∂u} \\ \frac{∂w}{∂v}=\frac{∂w}{∂x}\ \frac{∂x}{∂v}+\frac{∂w}{∂y}\ \frac{∂y}{∂v}\end{array}\)

    1. Calculate \(dy\text{/}dx\) if \(y\) is defined implicitly as a function of \(x\) via the equation \(3{x}^{2}-2xy+{y}^{2}+4x-6y-11=0.\) What is the equation of the tangent line to the graph of this curve at point \((2,1)?\)
    2. Calculate \(∂z\text{/}∂x\) and \(∂z\text{/}∂y,\) given \({x}^{2}{e}^{y}-yz{e}^{x}=0.\)
    Antwort aufdecken
    1. Set \(f(x,y)=3{x}^{2}-2xy+{y}^{2}+4x-6y-11=0,\) then calculate \({f}_{x}\) and \({f}_{y}\text{:}\) \(\begin{array}{l}{f}_{x}=6x-2y+4 \\ {f}_{y}=-2x+2y-6.\end{array}\)
      The derivative is given by
      \[\frac{dy}{dx}=-\frac{∂f\text{/}∂x}{∂f\text{/}∂y}=-\frac{6x-2y+4}{-2x+2y-6}=\frac{3x-y+2}{x-y+3}.\]
      The slope of the tangent line at point \((2,1)\) is given by
      \[{\frac{dy}{dx}|}_{(x,y)=(2,1)}=\frac{3(2)-1+2}{2-1+3}=\frac{7}{4}.\]
      To find an equation of the tangent line, we use the point-slope form ():
      \[\begin{array}{lll}y-{y}_{0} & = & m(x-{x}_{0}) \\ y-1 & = & \frac{7}{4}(x-2) \\ y & = & \frac{7}{4}x-\frac{7}{2}+1 \\ y & = & \frac{7}{4}x-\frac{5}{2}.\end{array}\]
    2. We have \(f(x,y,z)={x}^{2}{e}^{y}-yz{e}^{x}.\) Therefore,
      \[\begin{array}{l}\frac{∂f}{∂x}=2x{e}^{y}-yz{e}^{x} \\ \frac{∂f}{∂y}={x}^{2}{e}^{y}-z{e}^{x} \\ \frac{∂f}{∂z}=\text{-}y{e}^{x}.\end{array}\]
      Using ,
      \[\begin{array}{lllllllllllll}\begin{array}{ll}\frac{∂z}{∂x} & =-\frac{∂f\text{/}∂x}{∂f\text{/}∂z} \\ & =-\frac{2x{e}^{y}-yz{e}^{x}}{\text{-}y{e}^{x}} \\ & =\frac{2x{e}^{y}-yz{e}^{x}}{y{e}^{x}}\end{array} & & & \text{and} & & & \begin{array}{ll}\frac{∂z}{∂y} & =-\frac{∂f\text{/}∂y}{∂f\text{/}∂z} \\ & =-\frac{{x}^{2}{e}^{y}-z{e}^{x}}{\text{-}y{e}^{x}} \\ & =\frac{{x}^{2}{e}^{y}-z{e}^{x}}{y{e}^{x}}.\end{array}\end{array}\]

  10. Find \(dy\text{/}dx\) if \(y\) is defined implicitly as a function of \(x\) by the equation \({x}^{2}+xy-{y}^{2}+7x-3y-26=0.\) What is the equation of the tangent line to the graph of this curve at point \((3,-2)?\)

    Antwort aufdecken

    \(\frac{dy}{dx}={\frac{2x+y+7}{2y-x+3}|}_{(3,-2)}=\frac{2(3)+(-2)+7}{2(-2)-(3)+3}=-\frac{11}{4}\)
    Equation of the tangent line: \(y=-\frac{11}{4}x+\frac{25}{4}\)

  11. Let \(w(x,y,z)=xy\ \text{cos}\ z,\) where \(x=t,y={t}^{2},\) and \(z=\text{arcsin}\ t.\) Find \(\frac{dw}{dt}.\)

    Antwort aufdecken

    \(\frac{dw}{dt}=\frac{3{t}^{2}-4{t}^{2}}{\sqrt{1-{t}^{2}}}\)

  12. Let \(w(t,v)={e}^{tv}\) where \(t=r+s\) and \(v=rs.\) Find \(\frac{∂w}{∂r}\) and \(\frac{∂w}{∂s}.\)

  13. If \(w=5{x}^{2}+2{y}^{2},x=-3s+t,\) and \(y=s-4t,\) find \(\frac{∂w}{∂s}\) and \(\frac{∂w}{∂t}.\)

    Antwort aufdecken

    \(\frac{∂w}{∂s}=94s-46t,\) \(\frac{∂w}{∂t}=-46s+74t\)

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

  15. If \(f(x,y)=xy,x=r\ \text{cos}\ \theta ,\) and \(y=r\ \text{sin}\ \theta ,\) find \(\frac{∂f}{∂r}\) and express the answer in terms of \(r\) and \(\theta .\)

    Antwort aufdecken

    \(\frac{∂f}{∂r}=r\ \text{sin}(2\theta )\)

  16. Suppose \(f(x,y)=x+y,\) where \(x=r\ \text{cos}\ \theta\) and \(y=r\ \text{sin}\ \theta .\) Find \(\frac{∂f}{∂\theta }.\)

  17. \(f(x,y)={x}^{2}+{y}^{2},\) \(x=t,y={t}^{2}\)

    Antwort aufdecken

    \(\frac{df}{dt}=2t+4{t}^{3}\)

  18. \(f(x,y)=\sqrt{{x}^{2}+{y}^{2}},y={t}^{2},x=t\)

  19. \(f(x,y)=xy,x=1-\sqrt{t},y=1+\sqrt{t}\)

    Antwort aufdecken

    \(\frac{df}{dt}=-1\)

  20. \(f(x,y)=\frac{x}{y},x={e}^{t},y=2{e}^{t}\)

  21. \(f(x,y)=\text{ln}(x+y),\) \(x={e}^{t},y={e}^{t}\)

    Antwort aufdecken

    \(\frac{df}{dt}=1\)

  22. \(f(x,y)={x}^{4},\) \(x=t,y=t\)

  23. Let \(w(x,y,z)={x}^{2}+{y}^{2}+{z}^{2},\) \(x=\text{cos}\ t,y=\text{sin}\ t,\) and \(z={e}^{t}.\) Express \(w\) as a function of \(t\) and find \(\frac{dw}{dt}\) directly. Then, find \(\frac{dw}{dt}\) using the chain rule.

    Antwort aufdecken

    \(\frac{dw}{dt}=2{e}^{2t}\) in both cases

  24. Let \(z={x}^{2}y,\) where \(x={t}^{2}\) and \(y={t}^{3}.\) Find \(\frac{dz}{dt}.\)

  25. Let \(u={e}^{x}\text{sin}\ y,\) where \(x=-ln2t\) and \(y=\pi t.\) Find \(\frac{du}{dt}\) when \(x=\text{ln}\ 2\) and \(y=\frac{\pi }{4}.\)

    Antwort aufdecken

    \(\frac{du}{dt}=\sqrt{2}(\pi -4)\)

  26. \(\text{sin}(6x)+\text{tan}(8y)+5=0\)

  27. \({x}^{3}+{y}^{2}x-3=0\)

    Antwort aufdecken

    \(\frac{dy}{dx}=-\frac{3{x}^{2}+{y}^{2}}{2xy}\)

  28. \(\text{sin}(x+y)+\text{cos}(x-y)=4\)

  29. \({x}^{2}-2xy+{y}^{4}=4\)

    Antwort aufdecken

    \(\frac{dy}{dx}=\frac{y-x}{\text{-}x+2{y}^{3}}\)

  30. \(x{e}^{y}+y{e}^{x}-2{x}^{2}y=0\)

  31. \({x}^{2\text{/}3}+{y}^{2\text{/}3}={a}^{2\text{/}3}\)

    Antwort aufdecken

    \(\frac{dy}{dx}=\text{-}\sqrt[3]{\frac{y}{x}}\)

  32. \(x\ \text{cos}(xy)+y\ \text{cos}\ x=2\)

  33. \({e}^{xy}+y{e}^{y}=1\)

    Antwort aufdecken

    \(\frac{dy}{dx}=-\frac{y{e}^{xy}}{x{e}^{xy}+{e}^{y}(1+y)}\)

  34. \({x}^{2}{y}^{3}+\text{cos}\ y=0\)

  35. Find \(\frac{dz}{dt}\) using the chain rule where \(z=3{x}^{2}{y}^{3},x={t}^{4},\) and \(y={t}^{2}.\)

    Antwort aufdecken

    \(\frac{dz}{dt}=42{t}^{13}\)

  36. Let \(z=3\ \text{cos}\ x-\text{sin}(xy),x=\frac{1}{t},\) and \(y=3t.\) Find \(\frac{dz}{dt}.\)

  37. Let \(z={e}^{1-xy},x={t}^{1\text{/}3},\) and \(y={t}^{3}.\) Find \(\frac{dz}{dt}.\)

    Antwort aufdecken

    \(\frac{dz}{dt}=-\frac{10}{3}{t}^{7\text{/}3}\ \times \ {e}^{1-{t}^{10\text{/}3}}\)

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

  39. Let \(z=\frac{x}{y},x=2\ \text{cos}\ u,\) and \(y=3\ \text{sin}\ v.\) Find \(\frac{∂z}{∂u}\) and \(\frac{∂z}{∂v}.\)

    Antwort aufdecken

    \(\frac{∂z}{∂u}=\frac{-2\ \text{sin}\ u}{3\ \text{sin}\ v}\) and \(\frac{∂z}{∂v}=\frac{-2\ \text{cos}\ u\ \text{cos}\ v}{3\ {\text{sin}}^{2}v}\)

  40. Let \(z={e}^{{x}^{2}y},\) where \(x=\sqrt{uv}\) and \(y=\frac{1}{v}.\) Find \(\frac{∂z}{∂u}\) and \(\frac{∂z}{∂v}.\)

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Hier verwendete Symbole

Tippen Sie auf jedes Symbol für die vollständige Definition, ein Bild und was jeder Buchstabe darin bedeutet.

Wie man: The Chain Rule

  1. State the chain rules for one or two independent variables.
  2. Use tree diagrams as an aid to understanding the chain rule for several independent and intermediate variables.
  3. Perform implicit differentiation of a function of two or more variables.
  4. To use the chain rule, we need four quantities—
  5. To use the chain rule, we again need four quantities—
  6. Calculate
  7. Calculate
  8. Set

Fragen, die die Leute stellen

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen: the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

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Mehr in Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems