Q.If $x^m \cdot y^n = (x + y)^{m + n}$, prove that

Concept: Implicit Differentiation (logarithmic + implicit differentiation of a homogeneous relation).

Step 1 – Take the natural logarithm of both sides

Step 2 – Differentiate implicitly with respect to $x$

Step 3 – Solve for $\frac{dy}{dx}$

Multiply through by $xy(x+y)$ and collect terms:

After simplifying, the terms cancel to give: …

We start with the equation:

where $m$ and $n$ are constants. The exponents on both sides sum to the same total ($m+n$), which hints at homogeneity: if we scale $x$ and $y$ by a factor $t$, both sides scale by $t^{m+n}$. This symmetry often leads to a simple relationship between $y$ and $x$ — in fact, it suggests $y$ is proportional to $x$. Let’s verify that systematically.

1. Take natural logarithms on both sides to bring down the exponents:

Using logarithm properties:

This is now an implicit relation between $x$ and $y$, easier to differentiate.

2. Differentiate both sides with respect to $x$. Remember $y$ is a function of $x$, so $\frac{d}{dx}(\log y) = \frac{1}{y} \cdot \frac{dy}{dx}$.

This is the key equation we’ll solve for $\frac{dy}{dx}$.

3. Multiply through by $x y (x+y)$ to clear denominators (a clean algebraic move):

Expand carefully:

4. Collect terms with $\frac{dy}{dx}$ on one side and constants on the other:

Factor $\frac{dy}{dx}$ on the left:

5. Simplify the coefficients. Factor $x$ from the left bracket and $y$ from the right:

   Left: $x \left[ n(x+y) - (m+n) y \right] = x \left[ n x + n y - m y - n y \right] = x (n x - m y)$

   Right: $y \left[ (m+n) x - m (x+y) \right] = y \left[ m x + n x - m x - m y \right] = y (n x - m y)$

   So we have:

6. Assuming $n x - m y \neq 0$ (the non-degenerate case), we cancel this common factor: …

Method: Implicit + Logarithmic Differentiation for a Product/Power Relation

Use this method when $x$ and $y$ are tangled together inside products and powers (e.g. $x^m y^n = (x+y)^{m+n}$) — taking logs first turns awkward products and exponents into a sum that's far easier to differentiate implicitly.

Steps

Step 1: Take the natural logarithm of both sides

Use $\log(AB)=\log A+\log B$ and $\log(A^k)=k\log A$ to convert every product and exponent into a sum:

Step 2: Differentiate implicitly with respect to $x$

Remember $y$ is a function of $x$, so $\dfrac{d}{dx}(\log y) = \dfrac{1}{y}\dfrac{dy}{dx}$ (chain rule) — this extra $\dfrac{dy}{dx}$ factor appears on every $y$-term, including inside $\log(x+y)$.

Step 3: Clear denominators and collect $\dfrac{dy}{dx}$ terms …

Common Mistakes

Mistake 1: Forgetting the chain-rule factor $\dfrac{dy}{dx}$ on $\log y$

Why it's wrong: Differentiating $n\log y$ with respect to $x$ gives $\dfrac{n}{y}\cdot\dfrac{dy}{dx}$, not just $\dfrac{n}{y}$ — treating $y$ as if it were an independent variable like $x$ is the classic implicit-differentiation slip. Correct approach: every time $y$ appears inside a differentiated expression, multiply by $\dfrac{dy}{dx}$.

Mistake 2: Missing the chain-rule factor on $\log(x+y)$

Why it's wrong: Since $x+y$ contains $y$, differentiating $(m+n)\log(x+y)$ must produce $(m+n)\cdot\dfrac{1}{x+y}\cdot\left(1+\dfrac{dy}{dx}\right)$ — dropping the $\left(1+\dfrac{dy}{dx}\right)$ factor is an easy oversight because $x+y$ looks like a single variable. Correct approach: differentiate the inside of the log too, remembering it depends on both $x$ and $y$.

Mistake 3: Algebra errors collecting the $\dfrac{dy}{dx}$ terms …