Q.(a) If $x^m \cdot y^n=(x+y)^{m+n}$, then show that $\dfrac{dy}{dx}=\dfrac{y}{x}$

(a) $\log$: $m\ln x+n\ln y=(m+n)\ln(x+y)$; differentiate and simplify $\Rightarrow y'=\dfrac{y}{x}$. (b) Old income ₹540, new income ₹1437.50 $\Rightarrow$ change $=+\₹897.50$.

Part (a) — Implicit differentiation

Given: $x^m\cdot y^n=(x+y)^{m+n}$.

Step 1 — take logarithms of both sides.

$m\ln x+n\ln y=(m+n)\ln(x+y).$

Step 2 — differentiate with respect to $x$.

$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=(m+n)\cdot\frac{1+\dfrac{dy}{dx}}{x+y}.$

Step 3 — collect the $\dfrac{dy}{dx}$ terms.

$\frac{dy}{dx}\left(\frac{n}{y}-\frac{m+n}{x+y}\right)=\frac{m+n}{x+y}-\frac{m}{x}.$

The left bracket $=\dfrac{n(x+y)-y(m+n)}{y(x+y)}=\dfrac{nx-my}{y(x+y)}$, and the right side $=\dfrac{x(m+n)-m(x+y)}{x(x+y)}=\dfrac{nx-my}{x(x+y)}$.

Step 4 — solve for $\dfrac{dy}{dx}$.

$\frac{dy}{dx}=\frac{\dfrac{nx-my}{x(x+y)}}{\dfrac{nx-my}{y(x+y)}}=\frac{y(x+y)}{x(x+y)}=\frac{y}{x}.$

Part (b) — Stocks and shares

Step 1 — the first investment. Sundar buys ₹4,500 worth of 12% shares of face value ₹10 at par (market price $=$ face value $=$ ₹10).

$\text{Number of shares}=\frac{4500}{10}=450.$

$\text{Income}_1=12\%\text{ of face value}=0.12\times4500=\₹540.$

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