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Find \( \frac {dy}{dx} \) in the following: \( sin^2x +cos^2y = 1 \)

$\begin{array}{1 1}\large \frac{\sin\;2x}{\sin\;2y} \\ -\large \frac{\sin\;2x}{\sin\;2y} \\ \large \frac{\cos\;2x}{\cos\;2y} \\ -\large \frac{\cos\;2x}{\cos\;2y} \end{array} $

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Toolbox:
  • $\; \large \frac{d(sin^2x)}{dx} $$= 2\;sinx\;cosx = sin\;2x$
  • $\; \large \frac{d(cos^2x)}{dx} $$=- 2\;sinx\;cosx = -sin\;2x$
  • 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 $\large \frac{dy}{dx}$
Given $sin^2x+cos^2y=1$.
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 $\large \frac{dy}{dx}$
We know that $\; \large \frac{d(sin^2x)}{dx} $$= 2\;sinx\;cosx = sin\;2x$ and $\; \large \frac{d(cos^2x)}{dx} $$=- 2\;sinx\;cosx = -sin\;2x$.
Differentiating both sides:
$\Rightarrow sin \;2x\;dx - sin\;2y\;dy = 0$
$\Rightarrow \large \frac{dy}{dx} = \frac{\sin\;2x}{\sin\;2y}$
answered Apr 5, 2013 by balaji.thirumalai
 

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