The Chain Rule

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

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. \(z=f(x,y)=\sqrt{{x}^{2}-{y}^{2}},x=x(t)={e}^{2t},y=y(t)={e}^{\text{-}t}\)
   Жауап беріңіз

   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 \(t.\)

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

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

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

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

   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:

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

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

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

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

   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:

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

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

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 \(w.\)

   Жауап беріңіз

   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 \(w.\)

   Жауап беріңіз

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

9. 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 \({x}^{2}{e}^{y}-yz{e}^{x}=0.\)
   Жауап беріңіз

   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 \((3,-2)?\)

    Жауап беріңіз

    \(\frac{dy}{dx}={\frac{2x+y+7}{2y-x+3}|}_{(3,-2)}=\frac{2(3)+(-2)+7}{2(-2)-(3)+3}=-\frac{11}{4}\)
    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 \(\frac{dw}{dt}.\)

    Жауап беріңіз

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

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 \(\frac{∂w}{∂t}.\)

    Жауап беріңіз

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

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 \(\theta .\)

    Жауап беріңіз

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

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

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

    Жауап беріңіз

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

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

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

    Жауап беріңіз

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

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

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

    Жауап беріңіз

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

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.

    Жауап беріңіз

    \(\frac{dw}{dt}=2{e}^{2t}\) 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 \(y=\frac{\pi }{4}.\)

    Жауап беріңіз

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

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

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

    Жауап беріңіз

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

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

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

    Жауап беріңіз

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

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

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

    Жауап беріңіз

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

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

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

    Жауап беріңіз

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

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 \(y={t}^{2}.\)

    Жауап беріңіз

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

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 \(\frac{dz}{dt}.\)

    Жауап беріңіз

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

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 \(\frac{∂z}{∂v}.\)

    Жауап беріңіз

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

Өзіңіздіңіңізді сынап көріңіз

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Келесіде Multivariable Calculus

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