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  • If x^m.y^n = (x+y)^{m+n} prove that \frac{dy}{dx}= \frac{y}{x}

A.If x^m.y^n = (x+y)^{m+n}   prove that  \frac{dy}{dx}= \frac{y}{x}

B. If x^m.y^n = (x+y)^{m+n}   prove that  \frac{d^2y}{dx^2}=0

Answers (1)

Solution

 A. We have,

x^m.y^n = (x+y)^{m+n}

Taking log on both sides, we get

Taking log on both sides, we get
\\ \log \left(x^{m} y^{n}\right)=\log (x+y)^{m+n} \\\Rightarrow m \log x+n \log y=(m+n) \log (x+y)
Differentiating both sides w.r.t x, we get
\\ \mathrm{m} \cdot \frac{1}{\mathrm{x}}+\mathrm{n} \cdot \frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{m}+\mathrm{n}) \frac{1}{\mathrm{x}+\mathrm{y}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\mathrm{y}) \\=\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y} \cdot\left(1+\frac{d y}{d x}\right)

\\ \Rightarrow\left(\frac{n}{y}-\frac{m+n}{x+y}\right) \frac{d y}{d x}=\frac{m+n}{x+y}-\frac{m}{x} \\ \Rightarrow\left(\frac{n x+n y-m y-n y}{y(x+y)}\right) \frac{d y}{d x}=\left(\frac{m x+n x-m x-m y}{x(x+y)}\right) \\ \Rightarrow \quad\left(\frac{n x-m y}{y(x+y)}\right) \frac{d y}{d x}=\left(\frac{n x-m x}{x(x+y)}\right) \\ \Rightarrow \frac{d y}{d x}=\left(\frac{n x-m x}{x(x+y)}\right)\left(\frac{y(x+y)}{n x-m y}\right) \\ \Rightarrow \frac{d y}{d x}=\frac{y}{x}

Hence proved.

B. We have, 

\begin{aligned} &\frac{d y}{d x}=\frac{y}{x}\\ &\text { Differentiating both sides w.r.t } x, \text { we get }\\ &\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y} \cdot 1}{\mathrm{x}^{2}}\\ &\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}=\frac{\mathrm{x} \frac{\mathrm{y}}{\mathrm{x}}-\mathrm{y} \cdot \mathrm{.}}{\mathrm{x}^{2}}\left(\because \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}\right)\\ &\frac{d^{2} y}{d x^{2}}=\frac{y-y}{x^{2}}=0\\ &\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=0 \end{aligned}

Hence Proved

 

Posted by

infoexpert22

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