Question

Answer whichever question you can..... please mention the question number.....

class 12__ differentiation of implicit functions.​

Answer 1

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

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$\Large\fbox{\color{purple}{QUESTION}}$

$xy log(x + y) = 1$

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$\Large\fbox{\color{purple}{ SOLUTION }}$

ɴᴏᴡ,

ɴᴏᴡ,ᴅɪғғ, ᴡ.ʀ.ᴛ.x ᴡᴇ ɢᴇᴛ.

$\implies y log(x + y) + x \frac{dy}{dx} log(x + y) + ( \frac{xy}{x + y} )(1 + \frac{dy}{dx} ) = 0$

sɪɴᴄᴇ ,

$log(x + y) = \frac{1}{xy}$

ᴛʜᴇʀᴇғᴏʀᴇ,

$\implies \frac{y}{xy} + \frac{dy}{dx} ( \frac{x}{xy} ) + (\frac{xy}{x + y} )(1 + \frac{dy}{dx} ) = 0 \\ \implies \frac{1}{x} + \frac{1}{y} ( \frac{dy}{dx} ) + (\frac{xy}{x + y} ) + ( \frac{xy}{x + y} ) \frac{dy}{dx} = 0 \\ \implies \frac{dy}{dx} ( \frac{1}{y} + \frac{xy}{x + y} ) = - ( \frac{xy}{x + y} + \frac{1}{x} ) \\ \implies \frac{dy}{dx} ( \frac{x + y + x {y}^{2} }{y(x + y)}) = - ( \frac{ {x}^{2} y + x + y}{(x + y)x} ) \\ \implies \frac{dy}{dx} = - \frac{y}{x} \frac{( {x}^{2} y + x + y)}{(x + y + x {y}^{2} }$

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$\bf\Large\red{ THANKS \: FOR \: YOUR}$

$\bf\Large\red{ QUESTION \: HOPE \: IT }$

$\bf\Large\red{ HELPS }$

$\Large\mathcal\green{FOLLOW \: ME}$

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Answer 2

$hope \: it \: helps$