Question

# Find $$\displaystyle \frac{dy}{dx}$$ of $$\sin^2 y + \cos xy = k$$

Solution

## We have, $$\sin^2 y + \cos xy = k$$Differentiating both sides with respect to $$x$$, we obtain$$\Rightarrow \displaystyle \frac{d}{dx} (\sin^2 y) + \frac{d}{dx} (\cos xy) = \frac{d(\pi)}{dx}=0$$ .....(1)Using chain rule, we obtain$$\displaystyle \frac{d}{dx} (\sin^2 y) = 2 \sin y \frac{d}{dx} (\sin y) = 2 \sin y \cos y \frac{dy}{dx}$$...... (2)and $$\displaystyle \frac{d}{dx} (\cos xy) = - \sin xy \frac{d}{dx} (xy) = - \sin xy \left[ y \frac{d}{dx} (x) + x \frac{dy}{dx} \right]$$$$= - \sin xy \left[ y. 1 + x \dfrac{dy}{dx} \right] = -y \sin xy - x \sin xy \dfrac{dy}{dx}$$.....(3)From (1), (2) and (3), we obtain$$\displaystyle 2 \sin y \cos y \frac{dy}{dx} - y \sin xy - x \sin xy \frac{dy}{dx} = 0$$$$\Rightarrow \displaystyle (2 \sin y \cos y - x \sin xy) \frac{dy}{dx} = y \sin xy$$$$\Rightarrow \displaystyle (\sin 2 y - x \sin xy) \frac{dy}{dx} = y \sin xy$$$$\therefore \displaystyle \frac{dy}{dx} = \frac{y \sin xy}{\sin 2 y - x \sin xy}$$MathematicsRS AgarwalStandard XII

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