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

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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(sin^2x)}{dx} $$= 2\;sinx\;cosx = sin\;2x$
Given $sin^2y + cos \: xy = k$
This is not a standard differentiation of the form $y = f(x)$. We need to differentiate each term separately on LHS and RHS and then calculate $\large \frac{dy}{dx}$
$\textbf{Step 1}$:
Let us first calculate $\; \large \frac{d(\cos xy)}{dx} $, we'll apply the chain rule, with $g=xy$
According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
$\Rightarrow g' = x\;dy+y\;dx$
$\Rightarrow f'(g) = -\sin xy$
$\Rightarrow d(\cos xy) = f'(g).g' = -\sin xy. (x\;dy+y\;dx)$
$\textbf{Step 2}$:
We know that $\; \large \frac{d(sin^2y)}{dy} $$=\sin \;2y$
Therefore, if we differentiate both sides, we get:
$\Rightarrow \sin 2y \; dy -\sin xy. (x\;dy+y\;dx) = 0$
$\Rightarrow (\sin 2y \; -x\;\sin xy) \;dy = y\; \sin xy\;dx$
$\Rightarrow \large \frac{dy}{dx}= \frac{y\;\sin xy}{\sin 2y - x\;\sin\;xy}$

 

answered Apr 5, 2013 by balaji.thirumalai
 

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