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If $x^{m}\cdot y^{n}= \left ( x+y \right )^{m+n},$ then $dy/dx \: is$

Option 1)
$\frac{y}{x}$

Option 2)
$\frac{x+y}{xy}$

Option 3)
$xy$

Option 4)
$\frac{x}{y}$

Answers (1)

best_answer

As we learnt in

Chain Rule for differentiation (indirect) -

Let y = f(x) is not in standard form then

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

$ex:\:\:y=sin(ax+b)$

$Let\:\;u=(ax+b)$

$then\:\:y=sin \:u$

$so\:\:\frac{dy}{du}=cos \:u\:\:and\:\:\frac{du}{dx}=a$

$\therefore \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=a\:cos \:u$

$=a\:cos(ax+b)$

- wherein

$Where\;\:y=f(u)\:\;and\;\;u=f(x)$

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

$m\:log\:x +n\:log\:y =(m+n)log(x+y)$

$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{n+y}(1+\frac{dy}{dx})$

$\frac{m}{x}-\frac{m+n}{n+y}=\frac{dy}{dx}[\frac{m+n}{n+y}-\frac{n}{y}]$

$=\frac{mx+my-mn-nx}{x(n+y)}=\frac{dy}{dx}\left[\frac{my+ny-nx-ny}{y(x+y)} \right ]$

$\frac{dy}{dx}=\frac{y}{x}$

Option 1)

$\frac{y}{x}$

Correct

Option 2)

$\frac{x+y}{xy}$

Incorrect

Option 3)

$xy$

Incorrect

Option 4)

$\frac{x}{y}$

Incorrect

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