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in Sets, Relations and Functions by (15 points)

if x=(1+2x).e(y/x)

prove that: x3y"=(xy'-y)2

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1 Answer

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by (25.9k points)

x = (1 + 2x) ey/x 

Taking log both side, we get

log x = log(1 + 2x) + y/x log e

\(\Rightarrow\) log x = log(1 + 2x) + y/x (∵ log e = 1)

\(\Rightarrow\) y = x log x - x log (1 + 2x) .....(1)

Differentiating (1) w.r.t x, we get

y' = log x + x/x - log(1 + 2x) - 2x/(1 + 2x)

= log x - log(1 + 2x) + \(\frac{1 + 2x - 2x}{1 + 2x}\)

\(\Rightarrow\) y' = log x - log(1 + 2x) + \(\frac{1}{1 + 2x}\) ......(2)

Differentiating (2) w.r.t x we get

\(y'' = \frac{1}{x} - \frac{2}{1 + 2x} - \frac{2}{(1 + 2x)^2}\)

\(=\frac{(1 + 2x)^2 - 2(1 + 2x)x - 2x}{x(1 + 2x)^2}\) \(=\frac{4x^2 + 4x + 1 - 4x^2 - 2x - 2x}{x(1 + 2x)^2}\)

\(=\frac{1}{x(1 + 2x)^2}\)

Now xy' - y = x log \(\frac{x}{1 + 2x} + \frac{x}{1 + 2x} - x log \frac{x}{1 + 2x}\) = \(\frac{x}{1 + 2x}\)

∴ (xy' - y)2 = \(\frac{x^2}{(1 + 2x)^2}\) = x3 × \(\frac{1}{x(1 + 2x)^2}\) = x3y''

Hence, x3y'' = (xy' - y)2 Hence proved.

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