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

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

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:
$\Rightarrow sin \;2x\;dx - sin\;2y\;dy = 0$
$\Rightarrow \large \frac{dy}{dx} = \frac{\sin\;2x}{\sin\;2y}$