Question

If $\mathit{u}\mathbf{+}\mathit{v}\mathbf{=}{\mathit{x}}^{\mathbf{3}}\mathbf{-}{\mathit{y}}^{\mathbf{3}}\mathbf{+}\mathbf{4}\mathbf{,}{\mathit{u}}^{\mathbf{2}}\mathbf{-}{\mathit{v}}^{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$, find ${\left(\frac{\partial u}{\partial x}\right)}_{\mathbf{y}}\mathbf{,}{\left(\frac{\partial u}{\partial x}\right)}_{\mathbf{v}}\mathbf{,}{\left(\frac{\partial x}{\partial u}\right)}_{\mathbf{y}}\mathbf{,}{\left(\frac{\partial x}{\partial u}\right)}_{\mathbf{v}}$at$\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{u}\mathbf{,}\mathbf{v}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{)}$.

Short answer

The value of${\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}\right)}_{\mathrm{y}}$ is $\frac{4}{3}$.

The value of${\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}\right)}_{\mathrm{v}}$ is $\frac{14}{5}$.

The value of${\left(\frac{\partial \mathrm{x}}{\partial \mathrm{u}}\right)}_{\mathrm{y}}$ is $\frac{3}{4}$.

The value of${\left(\frac{\partial \mathrm{x}}{\partial \mathrm{u}}\right)}_{\mathrm{v}}$ is $\frac{5}{14}$.

Step-by-step solution

Given information.

Given$u^2+v^2=x^3-y^3+4$,$u^2-v^2=x^2{y}^2+1$

Definition of partial differentiation.

Partial differentiation is defined as the process, in which find the partial derivative of a function.

In Partial differentiation, the function has more than one variable and find the partial derivative of a function with respect to one variable and keeping the other variable constant.

Find (∂u∂x)y.

Find the partial differentiation of function,$u^2+v^2=x^3-y^3+4$with respect to x.

$2u\frac{\partial u}{\partial x}+2v\frac{\partial v}{\partial x}=3x^2-3y^2\frac{\partial y}{\partial x}...........\left(1\right)$

Now find the partial differentiation of function,$u^2-v^2=x^2{y}^2+1$with respect to x.

$2u\frac{\partial u}{\partial x}-2v\frac{\partial v}{\partial x}=2x{y}^2+2x^{2}y\frac{\partial y}{\partial x}........\left(2\right)$

Add both equations and simplify it.

$$\begin{gathered} 4u\frac{\partial u}{\partial x}=3x^2+2x{y}^2-3y^2\frac{\partial y}{\partial x}+2x^{2}y\frac{\partial y}{dx} \\ \frac{\partial u}{\partial x}=\frac{1}{u}\left[3x^2+2x{y}^2-3y^2\frac{\partial v}{\partial x}+2x^{2}y\frac{\partial y}{dx}\right] \end{gathered}$$

Now find${\left(\frac{\partial u}{\partial x}\right)}_y$ by keeping is constant.

$$\begin{gathered} {\left(\frac{\partial u}{\partial x}\right)}_y=\frac{1}{4.3}\left(3.2^2+2.2.{\left(-1\right)}^2\right) \\ =\frac{1}{12}\left(12+4\right) \\ =\frac{16}{12} \\ =\frac{4}{3} \end{gathered}$$

Substitute the values of ${\left(\frac{\partial u}{\partial x}\right)}_y$is $\frac{4}{3}$.

Find (∂u∂x)v. 

From equation (1), find ${\left(\frac{\partial \mathrm{u}}{\partial \mathrm{u}}\right)}_{\mathrm{v}}$ by keeping v is a constant.

$$\begin{gathered} 2u{\left(\frac{\partial \mathrm{u}}{\partial x}\right)}_{\mathrm{v}}+2v.0=3x^2-3y^2\frac{\partial y}{\partial x} \\ 2u{\left(\frac{\partial \mathrm{u}}{\partial x}\right)}_{\mathrm{v}}=3x^2-3y^2\frac{\partial y}{\partial x}...........\left(3\right) \end{gathered}$$

Similarly from equation (2), find ${\left(\frac{\partial u}{\partial x}\right)}_v$by keeping v is a constant.

$$\begin{gathered} 2u{\left(\frac{\partial y}{\partial x}\right)}_v-2y.0=2x{y}^2+2x^{2}y\frac{\partial y}{\partial x} \\ 2u{\left(\frac{\partial y}{\partial x}\right)}_v=2x{y}^2+2x^{2}y2u\frac{\partial y}{\partial x}...........\left(4\right) \end{gathered}$$

From equation (3) and (4),

$$\begin{gathered} 3x^2-3y^2\frac{\partial y}{\partial x}=2x{y}^2+2x^{2}y\frac{\partial y}{\partial x} \\ -\frac{\partial y}{\partial x}\left(3y^2+2x^{2}y\right)=2x{y}^2-3x^2 \\ \frac{\partial y}{\partial x}=\frac{3x^2-2x{y}^2}{3y^{2}2x^{2}y}...........\left(5\right) \end{gathered}$$

Now substitute the value of$\frac{\partial y}{\partial x}$from equation (5) into equation (3) and find ${\left(\frac{\partial u}{\partial x}\right)}_v$.

$$\begin{gathered} 2u{\left(\frac{\partial u}{\partial x}\right)}_v=3x^2-3y^2\left(\frac{3x^2-2x{y}^2}{3y^2+2x^{2}y}\right) \\ 2u{\left(\frac{\partial u}{\partial x}\right)}_v=\frac{9x^2{y}^2+6x^{2}y-9x^2{y}^2+6x{y}^4}{3y^2+2x^{2}y} \\ {\left(\frac{\partial u}{\partial x}\right)}_v=\frac{1}{2u}\left(\frac{9x^2{y}^2+6x^{2}y-9x^2{y}^2+6x{y}^4}{3y^2+2x^{2}y}\right) \end{gathered}$$

Substitute the values $(x,y,u,v)=(2,-1,3,2)$.

$$\begin{gathered} {\left(\frac{\partial u}{\partial x}\right)}_v=\frac{1}{2u}\left(\frac{9x^2{y}^2+6x^{2}y-9x^2{y}^2+6x{y}^2}{3y^2+2x^{2}y}\right) \\ =\frac{1}{6}\left(\frac{9\times 4\times 1-6\times 16\times 1-9\times 4\times 1+6\times 2\times 1}{3\times 1-2\times 4\times 1}\right) \\ =\frac{1}{6}\left(\frac{36-96-36+12}{3-8}\right) \\ =\frac{1}{6}\left(\frac{-84}{-5}\right) \\ =\frac{14}{5} \\ \mathrm{Hence}\mathrm{the}\mathrm{value}\mathrm{of}{\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}\right)}_{\mathrm{v}}\mathrm{is}\frac{14}{5}. \end{gathered}$$

Find (∂x∂u)y.

Find the partial differentiation of function,${\mathrm{u}}^2+{\mathrm{v}}^2={\mathrm{x}}^3-{\mathrm{y}}^3+4$with respect to u.

$2u+2v\frac{\partial v}{\partial u}=3x^2\frac{\partial x}{\partial u}-3y^2\frac{\partial y}{\partial u}..........\left(6\right)$

Now find the partial differentiation of function,${\mathrm{u}}^2-{\mathrm{v}}^2={\mathrm{x}}^2{\mathrm{y}}^2+1$ with respect to u.

$2u-2v\frac{\partial v}{\partial u}=2x{y}^2\frac{\partial x}{\partial u}+2x^{2}y\frac{\partial y}{\partial u}........\left(7\right)$

Now find ${\left(\frac{\partial x}{\partial u}\right)}_y$from equation (6) by keeping y is constant.

$$\begin{gathered} 2u+2v\frac{\partial v}{\partial u}=3x^2{\left(\frac{\partial v}{\partial u}\right)}_y-3y^2.0 \\ {\left(\frac{\partial x}{\partial u}\right)}_y=\frac{1}{3x^2}\left(2u+2v\frac{\partial v}{\partial u}\right).........\left(8\right) \end{gathered}$$

Now find${\left(\frac{\partial x}{\partial u}\right)}_y$ from equation (7) by keepingy is constant.

$$\begin{gathered} 2u-2v\frac{\partial v}{\partial u}=2x{y}^2{\left(\frac{\partial v}{\partial u}\right)}_y+2x{y}^2.0 \\ 2u-2v\frac{\partial v}{\partial u}=2x{y}^2{\left(\frac{\partial v}{\partial u}\right)}_y \\ {\left(\frac{\partial x}{\partial u}\right)}_y=\frac{1}{2x{y}^2}\left(2u-2v\frac{\partial v}{\partial u}\right).........\left(9\right) \end{gathered}$$

From equation (8) and (9), find the value of $\frac{\partial y}{\partial u}$.

$$\begin{gathered} \frac{1}{3x^2}\left(2u+2v\frac{\partial v}{\partial u}\right)=\frac{1}{2x{y}^2}\left(2u-2v\frac{\partial v}{\partial u}\right) \\ \left(2u+2v\frac{\partial v}{\partial u}\right)=\frac{3x^2}{2x{y}^2}\left(2u-2v\frac{\partial v}{\partial u}\right) \\ 2v\frac{\partial v}{\partial u}\left(1+\frac{3x}{2y^2}\right)=2u\left(\frac{3x}{2y^2}-1\right) \\ \frac{\partial v}{\partial u}=\left(\frac{u\left({\displaystyle \frac{3x}{2y^2}}-1\right)}{v\left(1+\frac{3x}{2y^2}\right)}\right) \end{gathered}$$

Now substitute the value of$\frac{\partial v}{\partial u}$in to equation (8).

$$\begin{gathered} {\left(\frac{\partial x}{\partial u}\right)}_y=\frac{1}{3x^2}\left(2u+2v.\frac{u\left({\displaystyle \frac{3x}{2y^2}}-1\right)}{v\left(1+\frac{3x}{2y^2}\right)}\right) \\ =\frac{1}{3x^2}\left(2u+2v.\frac{\left({\displaystyle \frac{3x}{2y^2}-1}\right)}{\left(1+\frac{3x}{2y^2}\right)}\right) \\ =\frac{2u}{3x^2}\left(1+\frac{\left({\displaystyle \frac{3x}{2y^2}-1}\right)}{\left(1+\frac{3x}{2y^2}\right)}\right) \end{gathered}$$

Substitute the values $(x,y,u,v)=(2,-1,3,2)$.

$$\begin{gathered} {\left(\frac{\partial x}{\partial v}\right)}_y=\frac{2u}{3x^2}\left(1+\frac{\left({\displaystyle \frac{3x}{2y^2}-1}\right)}{\left(1+\frac{3x}{2y^2}\right)}\right) \\ =\frac{6}{12}\left(1+\frac{\left({\displaystyle \frac{6}{2}-1}\right)}{\left(1+\frac{6}{2}\right)}\right) \\ =\frac{1}{2}\left(1+\frac{2}{4}\right) \\ =\frac{1}{2}\times \frac{3}{2} \\ =\frac{3}{2} \end{gathered}$$

Hence the value of${\left(\frac{\partial x}{\partial v}\right)}_y$is$\frac{3}{4}$.

Find (∂x∂u)v .

The partial differentiation of function,$u^2+v^2=x^3-y^3+4$ with respect tou is as follows:

$2u+2v\frac{\partial v}{\partial u}=3x^2\frac{\partial x}{\partial u}-3y^2\frac{\partial y}{\partial u}..........\left(10\right)$

The partial differentiation of function,$u^2-v^2=x^2{y}^2+1$ with respect to u. Is as follows:

$2u-2v\frac{\partial v}{\partial u}=2x{y}^2\frac{\partial x}{\partial u}+2x^{2}y\frac{\partial y}{\partial u}........\left(11\right)$

Now find${\left(\frac{\partial x}{\partial u}\right)}_v$ from equation (10) by keepingv is constant.

$$\begin{gathered} 2u+2v.0=3x^2{\left(\frac{\partial x}{\partial u}\right)}_v-3y^2\frac{\partial y}{\partial u} \\ 3x^2{\left(\frac{\partial x}{\partial u}\right)}_v=3y^2\frac{\partial y}{\partial u}+2u \\ {\left(\frac{\partial x}{\partial u}\right)}_v=\frac{1}{3x^2}\left(3x^2\frac{\partial y}{\partial u}+2u\right)............\left(12\right) \end{gathered}$$

Now find ${\left(\frac{\partial x}{\partial u}\right)}_v$ from equation (11) by keeping v is constant.

$$\begin{gathered} 2u+2v.0=2x{y}^2{\left(\frac{\partial x}{\partial u}\right)}_v+2x^{2}y\frac{\partial y}{\partial u} \\ 2x{y}^2{\left(\frac{\partial x}{\partial u}\right)}_v=2u-2x^{2}y\frac{\partial y}{\partial u} \\ {\left(\frac{\partial x}{\partial u}\right)}_v=\frac{1}{2x{y}^2}\left[2u-2x^{2}y\frac{\partial y}{\partial u}\right]............\left(13\right) \end{gathered}$$

From equation (12) and (13), find the value of $\frac{\partial y}{\partial u}$.

$$\begin{gathered} \frac{1}{3x^2}\left(3y^2\frac{\partial y}{\partial u}+2u\right)=\frac{1}{2x{y}^2}\left[2u-2x^{2}y\frac{\partial v}{\partial u}\right] \\ 3y^2\frac{\partial y}{\partial u}+2u=\frac{3x^2}{2x{y}^2}\left[2u-2x^{2}y\frac{\partial y}{\partial u}\right] \\ \frac{\partial y}{\partial u}\left(3y^2+\frac{6x^{2}y}{2x{y}^2}\right)=\frac{6x^{2}y}{2x{y}^2}-2u \\ \frac{\partial y}{\partial u}\left(3y^2+\frac{3x^2}{y}\right)=u\left(\frac{3x}{y^2}-2\right) \\ \frac{\partial y}{\partial u}\frac{u\left({\displaystyle \frac{3x}{y^2}}-2\right)}{\left(3y^2{\displaystyle \frac{3x^3}{y}}\right)} \end{gathered}$$

Now substitute the value of $\frac{\partial y}{\partial u}$in to equation (12).

$$\begin{gathered} {\left(\frac{\partial x}{\partial u}\right)}_v=\frac{1}{3x^2}\left(3y^2\frac{u\left({\displaystyle \frac{3x}{y^2}}-2\right)}{\left(3y^2+{\displaystyle \frac{3x^3}{y}}\right)}+2u\right) \\ =\frac{u{y}^2\left({\displaystyle \frac{3x}{y^2}}-2\right)}{x^2\left(3y^2{\displaystyle \frac{3x^3}{y}}\right)}+\frac{2u}{3x^2} \end{gathered}$$

Substitute the values $(x,y,u,v)=(2,-1,3,2)$.

$$\begin{gathered} {\left(\frac{\partial x}{\partial u}\right)}_v=\frac{u{y}^2\left({\displaystyle \frac{3x^2}{y^2}}-2\right)}{x^2\left(3y^2+{\displaystyle \frac{3x^3}{y}}\right)}+\frac{2u}{3x^2} \\ =\frac{3\left({\displaystyle \frac{6}{1}}-2\right)}{4\left(3+{\displaystyle \frac{24}{-1}}\right)}+\frac{6}{12} \end{gathered}$$

Now simplify the expression.

$$\begin{gathered} {\left(\frac{\partial x}{\partial u}\right)}_v=\frac{12}{-84}+\frac{1}{2} \\ =-\frac{1}{7}+\frac{1}{2} \\ =\frac{-2+7}{14} \\ =\frac{5}{14} \end{gathered}$$

Hence the value of${\left(\frac{\partial x}{\partial u}\right)}_v$ is $\frac{5}{14}$.