Q.$x$ and $y$ are the sides of two squares such that $y = x - x^2$. Find the rate of change of the area of the second square with respect to the area of the first square.

Idea: Write both areas as functions of $x$ and use $\dfrac{dA_2}{dA_1}=\dfrac{dA_2/dx}{dA_1/dx}$.

First square side $x$: $A_1=x^2$. Second square side $y=x-x^2$: $A_2=y^2=(x-x^2)^2$.

Differentiate each with respect to $x$:

Divide: …

The intuition

"Rate of change of $P$ with respect to $Q$" means the derivative $\frac{dP}{dQ}$. Here $P$ and $Q$ are the two areas, and both depend on the common variable $x$ (the side of the first square). When two quantities share a variable, we get $\frac{dP}{dQ}=\dfrac{dP/dx}{dQ/dx}$.

Set up

• First square: side $x$, so area $A_1=x^2$.
• Second square: side $y=x-x^2$, so area $A_2=y^2=(x-x^2)^2$.

We want $\dfrac{dA_2}{dA_1}$.

Work the steps

1. Differentiate $A_1$.

2. Differentiate $A_2$ (chain rule with $u=x-x^2$, $\frac{du}{dx}=1-2x$):

3. Divide to get $\frac{dA_2}{dA_1}$.

$\frac{dA_2}{dA_1}=\frac{dA_2/dx}{dA_1/dx}=\frac{2(x-x^2)(1-2x)}{2x}.$ …

Method: Finding the Rate of One Quantity With Respect to Another (Not Time) — the Quotient Trick

When a problem asks for $\frac{dP}{dQ}$ where neither $P$ nor $Q$ is time, but both are functions of a shared variable $x$, you don't need to introduce time at all — the chain rule gives a direct shortcut.

Steps

Step 1: Recognise the shared-variable structure.

Confirm both $P$ and $Q$ can be written explicitly as functions of the same variable $x$ (typically a length that determines both quantities).

Step 2: Write out $P(x)$ and $Q(x)$ using any given relation.

Substitute any relation connecting the underlying variables (e.g. one side expressed in terms of the other) so that both $P$ and $Q$ end up purely in terms of $x$.

Step 3: Differentiate both with respect to $x$ separately.

Compute $\frac{dP}{dx}$ and $\frac{dQ}{dx}$ as two ordinary derivatives, using the chain rule, product rule, etc. as each expression requires. …

Common Mistakes

Mistake 1: Trying to differentiate $A_2$ directly with respect to $A_1$ as if $A_1$ were an independent variable

Why it's wrong: neither area is given as an explicit function of the other; both are functions of the shared variable $x$, so $\frac{dA_2}{dA_1}$ must be computed via $\frac{dA_2/dx}{dA_1/dx}$, not by some direct differentiation of one area "with respect to" the other. Correct approach: differentiate both $A_1$ and $A_2$ with respect to $x$ separately, then divide.

Mistake 2: Forgetting the chain rule when differentiating $A_2=(x-x^2)^2$

Why it's wrong: writing $\frac{dA_2}{dx}=2(x-x^2)$ and stopping there omits the derivative of the inner function $(1-2x)$, which is required since $A_2$ is a composite function of $x$. Correct approach: $\frac{dA_2}{dx}=2(x-x^2)\cdot(1-2x)$. …