Question

If $x^m\cdot y^n=(x+y{)}^{m+n}$, then $\frac{dy}{dx}$ is ?

Answer 1

We have $x^m{y}^n={(x+y)}^{m+n}$

$\log x^m{y}^n=\log {(x+y)}^{m+n}$ by taking $\log$ both sides

$\log x^m+\log y^n=\log {(x+y)}^{m+n}$ since

$\log ab=\log a+\log b$

$m\log x+n\log y=(m+n)\log (x+y)$ since

$\log a^m=m\log a$

Differentiating both sides w.r.t $x$ we get

$m\frac{1}{x}+n\frac{1}{y}\frac{dy}{dx}=(m+n)\frac{1}{(x+y)}[1+\frac{dy}{dx}]$

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

$\Rightarrow \frac{mx+my-mx-nx}{x(x+y)}=\left(\frac{my+ny-nx-ny}{y(x+y)}\right)\frac{dy}{dx}$

$\Rightarrow \frac{my-nx}{x}=\left(\frac{my-nx}{y}\right)\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$