Question

if y = x/x+a prove that x dy/dx= y(1-y)

Answer 1

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

Given that,

$\rm :\longmapsto\:y = \dfrac{x}{x + a}$

can be rewritten as

$\rm :\longmapsto\:y = \dfrac{x + a - a}{x + a}$

$\rm :\longmapsto\:y = \dfrac{x + a}{x + a} - \dfrac{a}{x + a}$

$\rm :\longmapsto\:y = 1- \dfrac{a}{x + a}$

$\red{\rm :\longmapsto\:y = 1 - a {(x + a)}^{ - 1} \: }$

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

$\red{\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx} [1 - a {(x + a)}^{ - 1}] \: }$

$\red{\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx} 1 - a \dfrac{d}{dx} {(x + a)}^{ - 1} \: }$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} k \: = \: 0 \: }} \\ \\ and \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 0 + a {(x + a)}^{ - 2}\dfrac{d}{dx} (x + a)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = a {(x + a)}^{ - 2} (1+ 0)$

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} = \frac{a}{ {(x + a)}^{2} } \: }}$

Now, Consider

$\rm :\longmapsto\:y(1 - y)$

$\rm \:  =  \: \dfrac{a}{x + a} \bigg[1 - \dfrac{a}{x + a} \bigg]$

$\rm \:  =  \: \dfrac{a}{x + a} \bigg[\dfrac{x + a - a}{x + a} \bigg]$

$\rm \:  =  \: \dfrac{a}{x + a} \bigg[\dfrac{x}{x + a} \bigg]$

$\rm \:  =  \: x \times \dfrac{a}{ {(x + a)}^{2} }$

$\rm \:  =  \: x\dfrac{dy}{dx}$

Hence, Proved

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know :-

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