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Find \( \frac {dy}{dx} \) in the following: \( xy +y^2 = tan \: x + y \)

$\begin{array}{1 1} \frac{sec^2x-y}{x+2y-1} \\ \large \frac{sec^2x+y}{x+2y-1} \\ \frac{cosec^2x-y}{x+2y-1} \\ \frac{cosec^2x+y}{x+2y-1} \end{array} $

1 Answer

Toolbox:
  • For equations that are not of the form $y = f(x)$, we need to differentiate each term separately on LHS and RHS and then calculate $\large \frac{dy}{dx}$
  • $\; \large \frac{d(tanx)}{dx} $$= sec^2x$
Given $xy + y^2 = \tan x + y$
This is not a standard differentiation of the form $y = f(x)$. We need to differentiate each term separately on LHS and RHS and then calculate $\large \frac{dy}{dx}$
Differentiating both sides:
According to the Product Rule for differentiation, given two functions $u$ and $v, \large \frac {d(uv)}{dx} $$= u \large \frac{dv}{dx}$$+ v \large \frac{du}{dx}$
$\Rightarrow x\; dy + y\;dx + 2y \; dy = sec^2x \;dx + dy$
$\Rightarrow (x+2y-1)\; dy = (***^2x-y)\; dx$
$\large \frac{dy}{dx} = \frac{sec^2x-y}{x+2y-1}$
answered Apr 5, 2013 by balaji.thirumalai
 

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