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Find $\frac {dy}{dx}$ in the following: $ax +by^2 = cos \: y$

$\begin{array}{1 1} \large \frac{-a}{siny+2yb} \\ \large \frac{a}{siny+2yb} \\\large \frac{-a}{-siny+2yb} \\\large \frac{-a}{-siny+2yb} \end{array} $

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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(cosx)}{dx} $$=-sinx $
Given $ax+by^2 = \cos y$.
This is not a standard differentiation of the form $y = f(x)$. We need to differentiate each term separately on LHS and RHS and then calculate $\large \frac{dy}{dx}$
Differentiating both sides:
$\; \large \frac{d(cosx)}{dx} $$=-sinx $
$\Rightarrow a\; dx + b\; 2y \; dy = -siny \; dy$
$\Rightarrow -(siny+2yb)\; dy = a\;dx$
$\Rightarrow \large \frac{dy}{dx} = -\frac{a}{siny+2yb}$
answered Apr 5, 2013 by balaji.thirumalai
 

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