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If $x^my^n=(x+y)^{m+n}$,Prove that $(ii)\;\large\frac{d^2y}{dx^2}=$$0$

This is the second part of the multi-part question Q80

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Toolbox:
  • If $f'(x)$ is differentiable,we may differentiate this again w.r.t $x$,then the LHS becomes $\large\frac{d}{dx}\big(\large\frac{dy}{dx}\big)$ which is called the second order derivative .
Step 1:
Given : $x^{\large m}y^{\large n}=(x+y)^{\large m+n}$
Take $\log$ on both sides
$\log x^m+\log y^n=(m+n)\log (x+y)$
$\Rightarrow m\log x+n\log y=(m+n)\log (x+y)$
Differentiating w.r.t $x$ on both sides we get,
$m.\large\frac{1}{x}+$$n\large\frac{1}{y}\frac{dy}{dx}=$$(m+n).\large\frac{1}{(x+y)}$$\big(1+\large\frac{dy}{dx}\big)$
$\Rightarrow \large\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{(m+n)}{(x+y)}$$\big(1+\large\frac{dy}{dx}\big)$
Step 2:
$\large\frac{dy}{dx}\big(\large\frac{n}{y}-\frac{(m+n)}{(x+y)}\big)=\large\frac{(m+n)}{(x+y)}-\frac{m}{x}$
$\Rightarrow \large\frac{dy}{dx}\bigg(\large\frac{n(x+y)-y(m+n)}{y(x+y)}\bigg)=\large\frac{x(m+n)-m(x+y)}{x(x+y)}$
On simplifying we get,
$\Rightarrow \large\frac{dy}{dx}\big(\large\frac{nx-my}{y}\big)=\large\frac{nx-my}{x}$
Therefore $\large\frac{dy}{dx}=\frac{y}{x}$
Step 3:
Differentiating on both sides w.r.t $x$ we get,
(Apply quotient Rule)
$\large\frac{d^2y}{dx^2}=\large\frac{x.\Large\frac{dy}{dx}-\normalsize y.1}{x^2}$
$\Rightarrow \large\frac{x.\large\frac{dy}{dx}-y}{x^2}$
This can be written as
$\large\frac{1}{x}\frac{dy}{dx}-\large\frac{y}{x^2}$
But $\large\frac{dy}{dx}=\large\frac{y}{x}$
So substituting this we get,
$\large\frac{y}{x^2}-\frac{y}{x^2}$$=0$
Hence proved.
answered Jul 3, 2013 by sreemathi.v
 
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