Question

If xy=(x+y)^n and dy/dx= y/x, then n=

Answer 1

Step-by-step explanation:

We have,

$\tt{ \green{xy = (x + y)^{n}} }$

$\sf{ \implies \: y + x \dfrac{dy}{dx}= n(x + y)^{(n - 1)} \cdot \bigg(1 + \dfrac{dy}{dx} \bigg)}$

$\sf{ \implies \: y + x \dfrac{dy}{dx}= n(x + y)^{(n - 1)} + n(x + y)^{(n - 1)} \cdot \dfrac{dy}{dx} }$

$\sf{ \implies \: y - n(x + y)^{(n - 1)} = n(x + y)^{(n - 1)} \cdot \dfrac{dy}{dx} - x \dfrac{dy}{dx}}$

$\sf{ \implies \: y - n(x + y)^{(n - 1)} = \{n(x + y)^{(n - 1)} - x \} \dfrac{dy}{dx}}$

$\sf{ \implies \: \dfrac{y - n(x + y)^{(n - 1)} }{ n(x + y)^{(n - 1)} - x } = \dfrac{dy}{dx}}$

$\sf{ \implies \: \dfrac{dy}{dx} = \dfrac{y - n(x + y)^{(n - 1)} }{ n(x + y)^{(n - 1)} - x } }$

Now, it is given that $\rm{\dfrac{dy}{dx}=\dfrac{y}{x}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:xy = {(x + y)}^{n}$

On taking log on both sides, we get

$\rm :\longmapsto\:log(xy) = log {(x + y)}^{n}$

We know that,

$\boxed{ \tt{ \: logxy = logx + logy \: }}$

and

$\boxed{ \tt{ \: log {x}^{y} = ylogx \: }}$

So, using this, we get

$\rm :\longmapsto\:logx + logy = nlog(x + y)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}\bigg[logx + logy\bigg] = \dfrac{d}{dx}nlog(x + y)$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}logx = \frac{1}{x}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{d}{dx}logx + \dfrac{d}{dx}logy = n\dfrac{d}{dx}log(x + y)$

$\rm :\longmapsto\:\dfrac{1}{x} + \dfrac{1}{y}\dfrac{dy}{dx} = n\dfrac{1}{x + y} \dfrac{d}{dx}(x + y)$

$\rm :\longmapsto\:\dfrac{1}{x} + \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{n}{x + y}\bigg[1 + \dfrac{dy}{dx} \bigg]$

As it is given that,

$\boxed{ \tt{ \: \dfrac{dy}{dx} = \frac{y}{x} \: }}$

So, on substituting the value in above, we get

$\rm :\longmapsto\:\dfrac{1}{x} + \dfrac{1}{y} \times \dfrac{y}{x} = \dfrac{n}{x + y}\bigg[1 + \dfrac{y}{x} \bigg]$

$\rm :\longmapsto\:\dfrac{1}{x} + \dfrac{1}{x} = \dfrac{n}{x + y}\bigg[ \dfrac{x + y}{x} \bigg]$

$\rm :\longmapsto\:\dfrac{2}{x} = \dfrac{n}{x}$

$\rm \implies\:\boxed{ \tt{ \: \: n \: = \: 2 \: \: }}$

More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$