Practice Problems 11

Lecture 11: Partial Derivatives and Total Differentials

Problem numbers refer to the textbook (Chiang & Wainwright, 4th edition). These are not graded, but you should attempt them all before the quiz. Try each question yourself first, then expand the Solution to check.

Exercise 7.4

1. Find $\partial y/\partial x_1$ and $\partial y/\partial x_2$ for each of the following functions:

1. $y=2x_1^3-11x_1^2{x}_2+3x_2^2$

Solution

$$\frac{\partial y}{\partial x_1}=6x_1^2-22x_1{x}_2\frac{\partial y}{\partial x_2}=-11x_1^2+6x_2$$

4. $y={\displaystyle \frac{5x_1+3}{x_2-2}}$

Solution

$$\begin{array}{rl}\frac{\partial y}{\partial x_1}& =\frac{5(x_2-2)-(5x_1+3)\cdot 0}{{(x_2-2)}^2}=\frac{5}{x_2-2}\\ \frac{\partial y}{\partial x_2}& =\frac{0\cdot (x_2-2)-1\cdot (5x_1+3)}{{(x_2-2)}^2}=\frac{-(5x_1+3)}{{(x_2-2)}^2}\end{array}$$

2. Find $f_x$ and $f_y$ from the following:

1. $f(x,y)=x^2+5xy-y^3$

Solution

$$f_x=2x+5y{f}_y=5x-3y^2$$

2. $f(x,y)=(x^2-3y)(x-2)$

Solution

$$\begin{array}{rl}{f}_x& =(2x)(x-2)+(x^2-3y)\cdot 1=3x^2-4x-3y\\ f_y& =-3(x-2)+(x^2-3y)\cdot 0=-3x+6\end{array}$$

3. From the answers to Prob. 2, find $f_x(1,2)$, the value of the partial derivative $f_x$ when $x=1$ and $y=2$, for each function.

Solution

For (a), $f_x(1,2)=2(1)+5(2)=12$.

For (b), $f_x(1,2)=3(1{)}^2-4(1)-3(2)=-7$.

5. If the utility function of an individual takes the form $$U=U(x_1,x_2)={(x_1+2)}^2{(x_2+3)}^3$$ where $U$ is total utility, and $x_1$ and $x_2$ are the quantities of two commodities consumed:

1. Find the marginal-utility function of each of the two commodities.

Solution

$$\begin{array}{rl}{U}_1& =\frac{\partial U}{\partial x_1}=2(x_1+2){(x_2+3)}^3\\ U_2& =\frac{\partial U}{\partial x_2}=3{(x_1+2)}^2{(x_2+3)}^2\end{array}$$

2. Find the value of the marginal utility of the first commodity when 3 units of each commodity are consumed.

Solution

$$U_1(3,3)=2(3+2)(3+3{)}^3=2\times 5\times 216=2160$$

7. Write the gradients of the following functions:

1. $f(x,y,z)=x^2+y^3+z^4$

Solution

$$f_x=2x{f}_y=3y^2{f}_z=4z^3\mathrm{\nabla }f(x,y,z)=\left[\begin{array}{c}2x\\ 3y^2\\ 4z^3\end{array}\right]$$

2. $f(x,y,z)=xyz$

Solution

$$f_x=yz{f}_y=xz{f}_z=xy\mathrm{\nabla }f(x,y,z)=\left[\begin{array}{c}yz\\ xz\\ xy\end{array}\right]$$

Exercise 8.1

1. Find the differential $dy$, given:

1. $y=-x(x^2+3)$

Solution

Since $y=-x^3-3x$, $$dy=(-3x^2-3)dx=-3(x^2+1)dx$$

4. Find the point elasticity of demand, given $Q=k/P^n$, where $k$ and $n$ are positive constants.

1. Does the elasticity depend on the price in this case?

Solution

Writing $Q=k{P}^{-n}$, ${\displaystyle \frac{dQ}{dP}}=-nk{P}^{-n-1}$, so

$${\epsilon }_d=\frac{dQ}{dP}\cdot \frac{P}{Q}=\frac{-nk{P}^{-n-1}\cdot P}{k{P}^{-n}}=-n$$

No: the elasticity is the constant $-n$, whatever the price.

2. In the special case where $n=1$, what is the shape of the demand curve? What is the point elasticity of demand?

Solution

With $n=1$, $Q=k/P$, so $PQ=k$: the demand curve is a rectangular hyperbola, downward sloping and approaching both axes. The point elasticity is ${\epsilon }_d=-1$ everywhere (unit elastic).

6. Given $Q=100-2P+0.02Y$, where $Q$ is quantity demanded, $P$ is price, and $Y$ is income, and given $P=20$ and $Y=\mathrm{5,000}$, find the

1. Price elasticity of demand.

Solution

At $P=20$ and $Y=\mathrm{5,000}$, $Q=100-40+100=160$, so

$${\epsilon }_P=\frac{\partial Q}{\partial P}\cdot \frac{P}{Q}=-2\cdot \frac{20}{160}=-0.25$$

2. Income elasticity of demand.

Solution

$${\epsilon }_Y=\frac{\partial Q}{\partial Y}\cdot \frac{Y}{Q}=0.02\cdot \frac{5000}{160}=0.625$$

Exercise 8.2

3. Find the total differential, given

1. $y={\displaystyle \frac{x_1}{x_1+x_2}}$

Solution

$$\frac{\partial y}{\partial x_1}=\frac{1\cdot (x_1+x_2)-x_1\cdot 1}{{(x_1+x_2)}^2}=\frac{x_2}{{(x_1+x_2)}^2}$$

$$\frac{\partial y}{\partial x_2}=\frac{-x_1}{{(x_1+x_2)}^2}$$

$$dy=\frac{x_{2}d{x}_1-x_{1}d{x}_2}{{(x_1+x_2)}^2}$$

4. The supply function of a certain commodity is $$Q=a+b{P}^2+R^{1/2}(a<0,b>0)[R\text{: rainfall}]$$ Find the price elasticity of supply ${\epsilon }_{QP}$, and the rainfall elasticity of supply ${\epsilon }_{QR}$.

Solution

$${\epsilon }_{QP}=\frac{\partial Q}{\partial P}\cdot \frac{P}{Q}=\frac{2bP\cdot P}{Q}=\frac{2b{P}^2}{a+b{P}^2+R^{1/2}}$$

$${\epsilon }_{QR}=\frac{\partial Q}{\partial R}\cdot \frac{R}{Q}=\frac{1}{2}{R}^{-1/2}\cdot \frac{R}{Q}=\frac{R^{1/2}}{2(a+b{P}^2+R^{1/2})}$$

5. How do the two partial elasticities in Prob. 4 vary with $P$ and $R$? In a strictly monotonic fashion (assuming positive $P$ and $R$)?

Solution

Write $Q=a+b{P}^2+R^{1/2}$ and assume $Q>0$. For ${\epsilon }_{QP}=2b{P}^2/Q$:

$$\frac{\partial {\epsilon }_{QP}}{\partial P}=\frac{4bPQ-2b{P}^2\cdot 2bP}{Q^2}=\frac{4bP(a+R^{1/2})}{Q^2}$$

$$\frac{\partial {\epsilon }_{QP}}{\partial R}=\frac{-b{P}^2{R}^{-1/2}}{Q^2}<0$$

So ${\epsilon }_{QP}$ is strictly decreasing in $R$. In $P$ it is strictly increasing if $a+R^{1/2}>0$ and strictly decreasing if $a+R^{1/2}<0$: the sign does not depend on $P$, so it is strictly monotonic in $P$ either way.

For ${\epsilon }_{QR}=R^{1/2}/(2Q)$:

$$\frac{\partial {\epsilon }_{QR}}{\partial P}=\frac{-bP{R}^{1/2}}{Q^2}<0$$

$$\frac{\partial {\epsilon }_{QR}}{\partial R}=\frac{R^{-1/2}Q-1}{4Q^2}=\frac{R^{-1/2}(a+b{P}^2)}{4Q^2}$$

So ${\epsilon }_{QR}$ is strictly decreasing in $P$. In $R$ it is strictly increasing if $a+b{P}^2>0$ and strictly decreasing if $a+b{P}^2<0$: again the sign does not depend on $R$, so it is strictly monotonic in $R$.

6. The foreign demand for our exports $X$ depends on the foreign income $Y_f$ and our price level $P$: $X=Y_f^{1/2}+P^{-2}$. Find the partial elasticity of foreign demand for our exports with respect to our price level.

Solution

$${\epsilon }_{XP}=\frac{\partial X}{\partial P}\cdot \frac{P}{X}=\frac{-2P^{-3}\cdot P}{Y_f^{1/2}+P^{-2}}=\frac{-2}{Y_f^{1/2}{P}^2+1}<0$$

7. Find the total differential for each of the following functions:

2. $U=7x^2{y}^3$

Solution

$$\begin{array}{rl}dU& =U_{x}dx+U_{y}dy\\ & =14x{y}^{3}dx+21x^2{y}^{2}dy\\ & =7x{y}^2(2ydx+3xdy)\end{array}$$

6. $U=(x-3y{)}^3$

Solution

$$\begin{array}{rl}dU& =3(x-3y{)}^{2}dx+3(x-3y{)}^2(-3)dy\\ & =3(x-3y{)}^2(dx-3dy)\end{array}$$

Exercise 8.4

2. Find the total derivative $dz/dt$, given

1. $z=x^2-8xy-y^3$, where $x=3t$ and $y=1-t$

Solution

$$\begin{array}{rl}\frac{dz}{dt}& =z_x\frac{dx}{dt}+z_y\frac{dy}{dt}\\ & =(2x-8y)(3)+(-8x-3y^2)(-1)\\ & =14x-24y+3y^2\\ & =42t-24(1-t)+3(1-t{)}^2=3t^2+60t-21\end{array}$$

2. $z=7u+vt$, where $u=2t^2$ and $v=t+1$

Solution

$$\begin{array}{rl}\frac{dz}{dt}& =z_u\frac{du}{dt}+z_v\frac{dv}{dt}+z_t\\ & =7(4t)+t\cdot 1+v\\ & =28t+t+t+1=30t+1\end{array}$$

3. $z=f(x,y,t)$, where $x=a+bt$ and $y=c+kt$

Solution

$$\frac{dz}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}+f_t=b{f}_x+k{f}_y+f_t$$

4. Find the partial total derivatives $\mathrm{\S }W/\mathrm{\S }u$ and $\mathrm{\S }W/\mathrm{\S }v$ if

1. $W=a{x}^2+bxy+cu$, where $x=\alpha u+\beta v$ and $y=\gamma u$

Solution

$$\begin{array}{rl}\frac{\mathrm{\S }W}{\mathrm{\S }u}& =W_x\frac{\partial x}{\partial u}+W_y\frac{\partial y}{\partial u}+W_u\\ & =(2ax+by)\alpha +bx\gamma +c\\ & =(2a\alpha +b\gamma )(\alpha u+\beta v)+\alpha b\gamma u+c\end{array}$$

$$\begin{array}{rl}\frac{\mathrm{\S }W}{\mathrm{\S }v}& =W_x\frac{\partial x}{\partial v}+W_y\frac{\partial y}{\partial v}\\ & =(2ax+by)\beta +bx\cdot 0\\ & =\beta [(2a\alpha +b\gamma )u+2a\beta v]\end{array}$$

2. $W=f(x_1,x_2)$, where $x_1=5u^2+3v$ and $x_2=u-4v^3$

Solution

$$\frac{\mathrm{\S }W}{\mathrm{\S }u}=f_1\frac{\partial x_1}{\partial u}+f_2\frac{\partial x_2}{\partial u}=10u{f}_1+f_2$$

$$\frac{\mathrm{\S }W}{\mathrm{\S }v}=f_1\frac{\partial x_1}{\partial v}+f_2\frac{\partial x_2}{\partial v}=3f_1-12v^2{f}_2$$