Corrected exercises: implicit differentiation (201-103)

$2y\,y'=3x^{2}$, so $y'=\frac{3x^{2}}{2y}$. The point is on the curve: $3^{2}=2^{3}+1$. At $(2, 3)$: $y'=\frac{12}{6}=2$. Pitfall: differentiating $y^{2}$ as $2y$ without the factor $y'$, which leaves nothing to isolate.

$2x+2y\,y'=0$, so $y'=-\frac{x}{y}$. At $(2, -3)$: $y'=-\frac{2}{-3}=\frac{2}{3}\approx 0.667$. Pitfall: the sign. The ordinate is negative, so the slope is POSITIVE, which the picture of a circle in its lower right half confirms.

$18x+8y\,y'=0$, so $y'=-\frac{18x}{8y}=-\frac{9x}{4y}$. At $(2, 3)$: $y'=-\frac{18}{12}=-\frac{3}{2}=-1.5$. The coefficients of the ellipse carry over into the slope: $-\frac{9x}{4y}$, and not $-\frac{x}{y}$ as for a circle.

Product: $1\cdot y+x\,y'=0$, so $y'=-\frac{y}{x}$. At $(4, 5)$: $y'=-\frac{5}{4}=-1.25$. Pitfall: writing $(xy)'=y'$ and forgetting the term $y$, which gives $y'=0$.

Product, with a power of $y$: $y^{3}+x\cdot 3y^{2}y'=0$, so $y'=-\frac{y^{3}}{3xy^{2}}=-\frac{y}{3x}$. At $(2, 2)$: $y'=-\frac{2}{6}=-\frac{1}{3}\approx-0.333$. Pitfall: $(y^{3})'=3y^{2}$, without the $y'$.

Three terms, one of them a product: $2x+(y+x\,y')+2y\,y'=0$. Collect: $y'(x+2y)=-(2x+y)$, so $y'=-\frac{2x+y}{x+2y}$. At $(1, 2)$: $y'=-\frac{4}{5}=-0.8$. Pitfall: differentiating $xy$ as $y'$ alone, which gives $-\frac{2x}{1+2y}=-0.4$.

$\frac{1}{2\sqrt{x}}+\frac{y'}{2\sqrt{y}}=0$, so $y'=-\frac{\sqrt{y}}{\sqrt{x}}$. At $(1, 16)$: $y'=-\frac{4}{1}=-4$. The point satisfies the equation, $1+4=5$, and the square root of $y$ is differentiated like that of $x$, with its extra $y'$.

$3y^{2}y'+(y+x\,y')=0$, so $y'(3y^{2}+x)=-y$ and $y'=-\frac{y}{3y^{2}+x}$. At $(1, 2)$: $y'=-\frac{2}{12+1}=-\frac{2}{13}\approx-0.154$. The point satisfies the equation: $8+2=10$.

Take the logarithm of both sides, both positive at the point: $2\ln x+3\ln y=\ln 72$. Differentiate: $\frac{2}{x}+\frac{3y'}{y}=0$, so $y'=-\frac{2y}{3x}$. At $(3, 2)$: $y'=-\frac{4}{9}\approx-0.444$. The direct route gives the same thing, $2xy^{3}+3x^{2}y^{2}y'=0$, with more writing. Pitfall: $(\ln y)'=\frac{y'}{y}$, not $\frac{1}{y}$.

The chain rule on a DIFFERENCE: $3(x-y)^{2}\cdot(1-y')=1$, because the derivative of $x-y$ is $1-y'$. So $1-y'=\frac{1}{3(x-y)^{2}}$ and $y'=1-\frac{1}{3(x-y)^{2}}$. At $(8, 6)$, $x-y=2$: $y'=1-\frac{1}{12}=\frac{11}{12}\approx 0.917$. Pitfall: writing $3(x-y)^{2}$ without the factor $(1-y')$, which no longer contains any $y'$.

For $w=10$: $4\times 10-0.1\times 100=40-10=30$. For $w=30$: $4\times 30-0.1\times 900=120-90=30$. Both pairs work, so the abscissa $h=30$ carries TWO ordinates, $w=10$ and $w=30$. The vertical line test fails: the line $h=30$ cuts the curve twice, so $w$ is not a function of $h$. The figure shows it, the dashed vertical crosses the curve at two points. Economically, the reading is clear: thirty weekly hours can be obtained just as well with a modest wage as with a high one, for two opposite reasons.

Differentiate side by side with respect to $h$. On the left, $\dfrac{dh}{dh}=1$. On the right, each $w$ is an unknown function of $h$, so the chain rule puts a factor $\dfrac{dw}{dh}$ every time: $4\dfrac{dw}{dh}-0.2w\dfrac{dw}{dh}$. Hence $1=(4-0.2w)\dfrac{dw}{dh}$, then $\dfrac{dw}{dh}=\dfrac{1}{4-0.2w}$. The trap is to differentiate $0.1w^{2}$ as $0.2w$: that would amount to computing $\dfrac{dh}{dw}$, a different number, and making the equality false from the very first side.

At $w=10$: $\dfrac{dw}{dh}=\dfrac{1}{4-2}=0.5$ dollar per hour. At $w=30$: $\dfrac{dw}{dh}=\dfrac{1}{4-6}=-0.5$ dollar per hour. The two points have the SAME abscissa and opposite slopes, which is impossible for a function. The first sign says that you must pay fifty cents more to get one more hour, the usual substitution effect. The second says the opposite: at this wage level, getting one more hour requires LOWERING the wage by fifty cents, because the household, better paid, buys leisure time.

The derivative does not exist when the denominator vanishes, that is $4-0.2w=0$, so $w=20$ dollars, and then $h=4\times 20-0.1\times 400=80-40=40$ hours. The tangent there is VERTICAL, and this is not a quirk of the drawing: it is the maximum number of hours this household will ever supply. No change in wage, in either direction, buys the forty-first hour. Do not confuse this with a horizontal tangent, which would correspond to $\dfrac{dw}{dh}=0$ and never happens here.

Because implicit differentiation does not compute the slope of a function, it computes the slope of the CURVE at a POINT. You therefore have to give the point, not only the abscissa: at $(30, 10)$ the answer is $+0.50$ dollar, at $(30, 30)$ it is $-0.50$ dollar. That is exactly what the presence of $w$ in the expression $\dfrac{1}{4-0.2w}$ says, and it is why an implicit slope that depended on $h$ alone would be suspect.

$\dfrac{d}{dx}\bigl(5y^{2}\bigr)=10y\,\dfrac{dy}{dx}$, by the chain rule: differentiate the power, then multiply by the derivative of the inside, which is $\dfrac{dy}{dx}$. At $y=2$ and $\dfrac{dy}{dx}=3$: $10\times 2\times 3=60$. Next, $\dfrac{d}{dx}(xy)=y+x\,\dfrac{dy}{dx}$, the product rule; at $x=4$, $y=2$, $\dfrac{dy}{dx}=3$: $2+4\times 3=14$. Finally, $\dfrac{d}{dx}\bigl(y^{0.4}\bigr)=0.4\,y^{-0.6}\,\dfrac{dy}{dx}$; at $y=32$, since $32^{0.6}=8$, we have $32^{-0.6}=0.125$, so $0.4\times 0.125\times 3=0.15$.

Check: $2\times 15+3\times 16+0.05\times 15\times 16=30+48+12=90$, so the bundle is on the constraint. Differentiate: $2+3\dfrac{dy}{dx}+0.05\left(y+x\dfrac{dy}{dx}\right)=0$, the last term by the product rule. Collect: $2+0.05y+\left(3+0.05x\right)\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{2+0.05y}{3+0.05x}$. The right-hand side was 90, a constant, so its derivative is zero: it is this zero that makes the isolation possible.

At $(15, 16)$: the numerator equals $2+0.8=2.8$, the denominator $3+0.75=3.75$, so $\dfrac{dy}{dx}=-\dfrac{2.8}{3.75}\approx -0.747$. To buy one more unit of good A and stay within the budget, the household must give up $0.747$ unit of good B. The negative sign is the sign of a constraint: you cannot increase both quantities at the same time.

By forgetting the second term of the product rule, the student writes $2+0.05y+3\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{2.8}{3}\approx -0.933$. The ratio of the two values is $\dfrac{0.9333}{0.7467}=1.25$: the student overstates the sacrifice by exactly 25 percent. An error like this does not show to the naked eye, because the result stays negative and plausible, and that is what makes it costly on an exam.

Without the cross term, $2+3\dfrac{dy}{dx}=0$ gives $\dfrac{dy}{dx}=-\dfrac{2}{3}\approx -0.667$: the trade-off is simply the price ratio. The handling term adds $0.05y=0.80$ dollar to the marginal cost of good A, which is 40 percent of its price of 2 dollars, and $0.05x=0.75$ dollar to that of good B, which is 25 percent of its price of 3 dollars. Good A becomes proportionally more expensive, so the trade-off tips against it, from $0.667$ to $0.747$.

Differentiate both sides with respect to $x$. The term $xy$ is a product, so $y+x\,\dfrac{dy}{dx}$; the right-hand side is a constant, so zero. Hence $\dfrac{dy}{dx}=-\dfrac{y}{x}$. At $(4, 9)$, which is indeed on the curve since $4\times 9=36$, the slope is $-\dfrac{9}{4}=-2.25$. The household gives up $2.25$ units of the second good for one more unit of the first.

Here $y$ can be isolated: $y=\dfrac{36}{x}=36x^{-1}$, so $\dfrac{dy}{dx}=-36x^{-2}=-\dfrac{36}{x^{2}}$. At $x=4$: $-\dfrac{36}{16}=-2.25$, the same number. The two expressions coincide because, ON THE CURVE, $y=\dfrac{36}{x}$: substituting into the implicit form, $-\dfrac{y}{x}=-\dfrac{36/x}{x}=-\dfrac{36}{x^{2}}$. They do not coincide elsewhere, and that is normal: the implicit form only makes sense at a point of the curve.

Check: $3\times 9\times 1+1^{3}=27+1=28$. Differentiate: $3\left(y+x\dfrac{dy}{dx}\right)+3y^{2}\dfrac{dy}{dx}=0$, the first bracket by the product rule and the second term by the chain rule. Collect: $3y+\left(3x+3y^{2}\right)\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{y}{x+y^{2}}$. At $(9, 1)$: $-\dfrac{1}{9+1}=-0.1$. Here the explicit route is closed: isolating $y$ would require solving a cubic equation in $y$ for every $x$, and the Cardano formula that exists is of no practical use.

The slope predicts $\Delta y\approx -0.1\times 0.1=-0.010$. The actual change is $0.9901-1=-0.0099$. The gap between the two is $0.0001$, one percent of the change: the slope is confirmed. This check is the only honest way to test an implicit derivative when no explicit formula lets you recompute.

The rule: if $y$ can be isolated cleanly, both routes work and the explicit one gives a slope written as a function of $x$ ALONE; otherwise, the implicit route is the only one. What the implicit route gives in addition is that it works on curves that are the graph of no function, and that it delivers the slope as a function of the POINT, which is the useful form in economics, where the current bundle is always known.

Check: $12^{2}+4\times 8^{2}=144+256=400$. Differentiate both sides with respect to $x$: $2x+8y\,\dfrac{dy}{dx}=0$, the second term by the chain rule applied to $4y^{2}$. Hence $\dfrac{dy}{dx}=-\dfrac{2x}{8y}=-\dfrac{x}{4y}$. Writing $2x+8y=0$ would be the classic error, and here it would give $y=-3$, a negative production: the absurdity of the result gives away the forgotten factor.

$\text{MRT}=-\dfrac{dy}{dx}=\dfrac{12}{4\times 8}=\dfrac{12}{32}=0.375$. To produce one more thousand tonnes of agricultural goods, this economy must give up 375 units of industrial goods. The MRT is positive because the slope is negative: it is a sacrifice, and a sacrifice is always stated as a positive value.

At $(16, 6)$: $\text{MRT}=\dfrac{16}{24}=\dfrac{2}{3}\approx 0.667$, almost double the previous one. At $(0, 10)$: $\text{MRT}=0$, the tangent is horizontal, and that makes sense, since the first tonne of agricultural goods costs almost nothing when the economy produces none. At $(20, 0)$ the denominator $4y$ vanishes: the derivative does not exist, the tangent is VERTICAL. No quantity of industrial goods pays for the extra tonne, the economy is already fully specialized.

On the useful part, $y=\dfrac{\sqrt{400-x^{2}}}{2}$, so $\text{MRT}=\dfrac{x}{4y}=\dfrac{x}{2\sqrt{400-x^{2}}}$. As $x$ grows from 0 to 20, the numerator grows and the denominator shrinks, so the quotient is strictly increasing. This is the LAW OF INCREASING OPPORTUNITY COSTS, proved and not merely asserted: the more an economy specializes, the more each additional unit costs it, because it moves less and less suitable resources toward agriculture.

The tangent passes through $(12, 8)$ with slope $-0.375$: $y-8=-0.375(x-12)$, so $y=12.5-0.375x$. Its vertical intercept is $12.5$, whereas the maximum industrial production is $y=10$, reached at $x=0$. The tangent therefore promises $12.5$ thousand units where the frontier only gives 10. The MRT is a LOCAL rate: it holds in the neighbourhood of the point and cannot be extrapolated, just as a marginal cost is not multiplied by a hundred units.