Question

If x3y4=(x+y)n+1 and dy/dx=y/x then n=.,.​

Answer 1

I hope it will help you 3

Answer 2

Given,

$\[\begin{array}{l}{x^3}{y^4} = {\left( {x + y} \right)^{n + 1}} - - - - - \left( 1 \right)\\\frac{{dy}}{{dx}} = \frac{y}{x} - - - - - - \left( 2 \right)\end{array}\]$

Solution,

$\[{x^3}{y^4} = {\left( {x + y} \right)^{n + 1}}\]$

Taking log on both side,

$\[\begin{array}{l} \Rightarrow \log \left( {{x^3}{y^4}} \right) = \log {\left( {x + y} \right)^{n + 1}}\\ \Rightarrow 3\log x + 4\log y = \left( {n + 1} \right)\log \left( {x + y} \right)\end{array}\]$

Differentiate with respect to x,

$\[ \Rightarrow \frac{3}{x} + \frac{4}{y}\frac{{dy}}{{dx}} = \frac{{\left( {n + 1} \right)}}{{x + y}}\left( {1 + \frac{{dy}}{{dx}}} \right)\]$

$\[ \Rightarrow \frac{3}{x} + \frac{4}{y}\left( {\frac{y}{x}} \right) = \frac{{\left( {n + 1} \right)}}{{x\left( {1 + \frac{y}{x}} \right)}}\left( {1 + \frac{y}{x}} \right)\]$

$\[ \Rightarrow \frac{3}{x} + \frac{4}{x} = \frac{{\left( {n + 1} \right)}}{x}\]$

$\[ \Rightarrow \frac{7}{x} = \frac{{\left( {n + 1} \right)}}{x}\]$

$\[\begin{array}{l} \Rightarrow n + 1 = 7\\ \Rightarrow n = 6\end{array}\]$

Hence the value of n is 6.