Chapter 8 — Key Takeaways

A structured recap of implicit differentiation and related rates. Every idea in this chapter is the chain rule applied through a hidden dependency — either $y$ depending on $x$, or every quantity depending on time $t$. Hold onto that single sentence and the rest follows.


1. The Method of Implicit Differentiation (§8.2)

When a relation $F(x, y) = 0$ defines $y$ as a hidden function of $x$, find $\dfrac{dy}{dx}$ without solving for $y$:

  1. Differentiate both sides with respect to $x$.
  2. Apply the chain rule through $y$: every term containing $y$ produces a factor of $\dfrac{dy}{dx}$. A term $y^3$ differentiates to $3y^2\,y'$, not $3y^2$.
  3. Solve algebraically for $\dfrac{dy}{dx}$: collect the $y'$ terms, factor, divide.

The answer is a function of both $x$ and $y$ — and that is exactly the right amount of information, because a single point on the curve already knows which branch it is on.

Standard results worth recognizing

Relation $dy/dx$ Note
$x^2 + y^2 = c$ $-x/y$ circle; tangent $\perp$ radius (§8.2)
$xy = c$ $-y/x$ rectangular hyperbola
$x^3 + y^3 = 6xy$ $\dfrac{2y - x^2}{y^2 - 2x}$ folium of Descartes (§8.2)
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $-\dfrac{b^2 x}{a^2 y}$ ellipse

2. Tangents, Vertical Tangents, Second Derivatives

  • Tangent line at $(x_0, y_0)$: $\;y - y_0 = y'(x_0, y_0)\,(x - x_0)$ (§8.4).
  • Vertical tangents occur where the denominator of $y'$ vanishes while the numerator does not — features the explicit form $y = \pm\sqrt{\cdots}$ hides (§8.4).
  • Higher derivatives (§8.3): differentiate $y'$ again, treating $y$ as a function of $x$ and substituting the known $y'$ back in. For the unit circle this yields $y'' = -1/y^3$; the sign reads off concavity directly.

3. Inverse-Function Derivatives (§8.5)

The cleanest route to any inverse derivative. From $y = f^{-1}(x) \Leftrightarrow x = f(y)$, differentiate implicitly:

$$\boxed{(f^{-1})'(x) = \frac{1}{f'\!\big(f^{-1}(x)\big)}}$$

This one identity generates the catalog:

$$\frac{d}{dx}\arctan x = \frac{1}{1+x^2}, \qquad \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\ln x = \frac{1}{x}.$$

Sign caution: when a square root appears, the inverse function's range dictates the sign. For $\arcsin$, the range $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ forces $\cos y \ge 0$, so the positive root is correct.


4. Logarithmic Differentiation (§8.6)

When $y$ is a tangle of products, quotients, and powers — or has a variable in both base and exponent — take $\ln$ first, then differentiate implicitly:

  1. $\ln y = \ln f(x)$.
  2. Differentiate: $\dfrac{y'}{y} = \dfrac{d}{dx}[\ln f(x)]$ (logs turn products into sums, powers into multipliers).
  3. $y' = y\cdot\dfrac{d}{dx}[\ln f(x)]$.

Signature example: $y = x^x \Rightarrow y' = x^x(\ln x + 1)$ — there is no honest way to get this without the log trick.


A related-rates problem is implicit differentiation with respect to time. Multiple quantities change together; a connecting equation binds them; differentiating it with respect to $t$ relates all their rates at once.

  1. Read and identify — what is changing, what rate is given, what rate is wanted.
  2. Draw and label — assign letters to changing quantities, not numbers.
  3. Record the given and wanted rates in symbols ($\dfrac{dx}{dt}$, $\dfrac{dy}{dt}$, …).
  4. Find the relating equation — Pythagoras, similar triangles, a volume formula, a trig ratio.
  5. Differentiate both sides with respect to $t$ — chain rule on every variable.
  6. Substitute the known values for the chosen instant and the known rates.
  7. Solve for the unknown rate, then check units and sign.

The golden rule: the geometry holds always; the numbers hold only now. Differentiate while everything is still a variable; freeze the numbers only at step 6.

Setup Relating equation Worked in Field
Sliding ladder $x^2 + y^2 = L^2$ §8.8 safety/engineering
Conical tank filling $V = \tfrac13\pi r^2 h$, $r = kh$ §8.9 chemical/civil eng.
Approaching/separating cars $D^2 = x^2 + y^2$ §8.10 navigation, tracking
Walking shadow similar triangles, $s = \tfrac12 x$ §8.11 optics/lighting
Filling trough $V = (\text{cross-section})\times L$ §8.12 construction
Changing angle $\tan\theta = h/d$ §8.13 radar/camera tracking
Inflating sphere $V = \tfrac43\pi r^3$, $S = 4\pi r^2$ §8.14 balloons, imaging, biology

Key identity from the sphere: $\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}$ — volume rate = surface area × radial rate (the shell, a glimpse of Chapter 18's integration).


6. Implicit Differentiation Beyond Geometry (§8.15)

A held-constant constraint binds variables implicitly. For a Cobb–Douglas isoquant $A\,L^\alpha K^\beta = Q_0$, implicit differentiation gives the marginal rate of technical substitution:

$$\frac{dK}{dL} = -\frac{\alpha K}{\beta L}.$$

Same gesture as the circle's $-x/y$ — a constraint differentiated implicitly. Physics relates state variables $(P, V, T)$ via $PV = nRT$ the same way.


7. The Two Most Common Errors

  • Forgetting the chain-rule factor. Differentiating $y^3$ as $3y^2$ instead of $3y^2\,y'$. Tell-tale symptom: your answer for $y'$ contains no $y$, or $y'$ never appears in the differentiated equation. Every $y$-term must produce a $y'$.
  • Substituting numbers before differentiating (related rates). Plugging in $x = 6$ early turns $x$ into a constant with $\dfrac{dx}{dt} = 0$, throwing away the motion. Differentiate first; substitute at step 6.

Secondary traps: dropping a held-constant quantity entirely (its derivative is zero, but its value still appears in the equation); sign/quadrant errors in inverse-trig square roots; forgetting to eliminate a second changing length (e.g. $r$ in the cone) via similar triangles before differentiating.


8. Skills Checklist

  • [ ] Differentiate an implicit relation for $\dfrac{dy}{dx}$.
  • [ ] Find tangent and vertical-tangent lines to implicit curves.
  • [ ] Compute $y''$ implicitly and read off concavity.
  • [ ] Derive inverse-trig and $\ln$ derivatives from implicit differentiation.
  • [ ] Apply logarithmic differentiation to products/quotients/variable exponents.
  • [ ] Execute the seven-step related-rates method on ladders, tanks, vehicles, shadows, and tracking angles.
  • [ ] Check units and signs and interpret a negative rate physically.

9. Connections

  • Chapter 7 — the chain, product, and quotient rules used in every computation here.
  • Chapter 9 — curve analysis uses implicit $y'$ and $y''$ to locate tangents and concavity on non-function curves.
  • Chapter 10 — optimization often differentiates a constraint implicitly to eliminate a variable.
  • Chapter 19 — differential equations frequently produce solutions defined only implicitly, $F(x,y) = C$.
  • Chapter 31 — optimization in several variables: the isoquant/Lagrange-multiplier formalism is the multivariable successor to §8.15's MRTS.
  • Chapter 30 — the Implicit Function Theorem (§8.4 sidebar) and the gradient generalize implicit differentiation to many variables.

What's Next

Chapter 9 turns derivatives loose on the shape of functions — where they rise and fall, where they bend, where they peak — and several of those analyses lean on the implicit $y'$ and $y''$ you computed here. Chapter 10 then uses derivatives to find best-possible values, repeatedly calling on the implicit differentiation of constraints you practiced in §8.15.

Reflection

We did not invent a new kind of derivative. Implicit differentiation and related rates are the Chapter 7 chain rule, applied with the willingness to differentiate an equation we cannot solve and to let the answer involve more than one variable. A circle's tangent falls out as $-x/y$, a ladder's fall as $-\tfrac{x}{y}\dfrac{dx}{dt}$, an economy's substitution rate as $-\dfrac{\alpha K}{\beta L}$ — all the same gesture, aimed at different worlds. Calculus is the mathematics of change, and change almost never arrives pre-solved for $y$. Now you can differentiate it anyway.