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

(i) $\frac{{dy}}{{dx}} = \frac{y}{x}$ and

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

We have, ${x^m} \cdot {y^n} = {(x + y)^{m + n}}$ ………(i)

(i) Differentiating Eq. (i) w.r.t. $x$, we get
$\frac{d}{{dx}}\left( {{x^m} \cdot {y^n}} \right) = \frac{d}{{dx}}{(x + y)^{m + n}}$

$\Rightarrow$ ${x^m} \cdot \frac{d}{{dy}}{y^n} \cdot \frac{{dy}}{{dx}} + {y^n} \cdot \frac{d}{{dx}}{x^m} = (m + n){(x + y)^{m + n - 1}}\frac{d}{{dx}}(x + y)$

$\Rightarrow$ ${x^m} \cdot n{y^{n - 1}}\frac{{dy}}{{dx}} + {y^n} \cdot m{x^{m - 1}} = (m + n){(x + y)^{m + n - 1}}\left( {1 + \frac{{dy}}{{dx}}} \right)$

$\Rightarrow$ $\frac{{dy}}{{dx}}\left[ {{x^m} \cdot n{y^n}^{ - 1} - (m + n) \cdot {{(x + y)}^{m + n - 1}}} \right] = (m + n){(x + y)^{m + n - 1}} - {y^n}m{x^{m - 1}}$

$\Rightarrow$ $\frac{{dy}}{{dx}}\left[ {n{x^m}{y^{n - 1}} - (m + n){{(x + y)}^{m + n - 1}}} \right] = (m + n) \cdot {(x + y)^{m + n - 1}} - \frac{{{y^n} - 1 \cdot y \cdot m{x^m}}}{x}$

therefore,

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

$= \frac{{\frac{{x(m + n){{(x + y)}^{m + n}} - (x + y) \cdot y{ \cdot ^{n - 1}}y \cdot m{x^m}}}{{(x + y) \cdot x}}}}{{\frac{{(x + y)n{x^m}{y^n} - y(m + n){{(x + y)}^{m + n}}}}{{(x + y) \cdot y}}}}$

$= \frac{{\frac{{x(m + n) \cdot {x^m} \cdot {y^n} - m(x + y){y^n}{x^m}}}{{(x + y) \cdot x}}}}{{\frac{{(x + y)n{x^m} \cdot {y^n} - y(m + n) \cdot {x^m} \cdot {y^n}}}{{(x + y) \cdot y}}}}$

$= \frac{{{x^m}{y^n}[mx + nx - mx - my] \cdot (x + y)y}}{{{x^m}{y^n}[nx + ny - my - ny] \cdot (x + y) \cdot x}}$

$= \frac{y}{x}$ ……(ii)

Hence proved.

(ii) Further, differentiating Eq.(ii) i.e., $\frac{{dy}}{{dx}} = \frac{y}{x}$ on both the sides w.r.t. $x$, we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x \cdot \frac{{dy}}{{dx}} - y \cdot 1}}{{{x^2}}}$

$= \frac{{x \cdot \frac{y}{x} - y}}{{{x^2}}}$

$= 0$
Hence proved.