Question

If $x^m\cdot y^n=(x+y{)}^{m+n}$, then which of the following is/are true :

• A. $\frac{dy}{dx}=\frac{y}{x}$
• B. $\frac{dy}{dx}=-\frac{y}{x}$
• C. $\frac{d^{2}y}{d{x}^2}=\frac{x^2+y^2}{x^2}$
• D. $\frac{d^{2}y}{d{x}^2}=0$

Answer 1

Answer: (A), (D)

$\frac{d^{2}y}{d{x}^2}=0$

Given $x^m\cdot y^n=(x+y{)}^{m+n}$
Applying $\ln$ both the sides
$m\ln \left(x\right)+n\ln \left(y\right)=(m+n)\ln (x+y)$
Differentiating w.r.t $x$
$\Rightarrow \frac{m}{x}+\frac{n}{y}{y}^{\prime }=\frac{m+n}{x+y}(1+y^{\prime })$
$\Rightarrow y^{\prime }(\frac{n}{y}-\frac{m+n}{x+y})=\frac{m+n}{x+y}-\frac{m}{x}$
$\Rightarrow \frac{y^{\prime }(nx-my)}{y(x+y)}=\frac{(nx-my)}{x(x+y)}$
$\Rightarrow y^{\prime }=\frac{y}{x}\cdots \left(i\right)$
Differenting w.r.t $x$
$\Rightarrow y^{\prime \prime }=\frac{x{y}^{\prime }-y}{x^2}=\frac{x\left(\frac{y}{x}\right)-y}{x^2}=0$ from $\left(i\right)$
$\therefore y^{\prime \prime }=0$