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If $x^my^n=(x+y)^{m+n}$,Prove that $(ii)\;\large\frac{d^2y}{dx^2}=$$0 This is the second part of the multi-part question Q80 Can you answer this question? 1 Answer 0 votes Toolbox: • If f'(x) is differentiable,we may differentiate this again w.r.t x,then the LHS becomes \large\frac{d}{dx}\big(\large\frac{dy}{dx}\big) which is called the second order derivative . Step 1: Given : x^{\large m}y^{\large n}=(x+y)^{\large m+n} Take \log on both sides \log x^m+\log y^n=(m+n)\log (x+y) \Rightarrow m\log x+n\log y=(m+n)\log (x+y) Differentiating w.r.t x on both sides we get, m.\large\frac{1}{x}+$$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.