Question

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$find \: derivative \: of \: \frac{dy}{dx} \: of \\ { \sin}^{2} y \: + \cos(xy)$
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Answer 1

Step-by-step explanation:

We have,

$\mapsto{ \purple{\bf{y = sin^{2} (y) + cos(xy)}}}$

$\implies \sf{ \dfrac{dy}{dx}= 2 \: sin (y) \: cos(y) \: \dfrac{dy}{dx} - sin(xy) \cdot \left \{ y + x \dfrac{dy}{dx} \right \}}$

$\implies \sf{ \dfrac{dy}{dx}= sin (2y) \: \dfrac{dy}{dx} - y \: sin(xy) - x \: sin(xy)\dfrac{dy}{dx} }$

$\implies \sf{ y \: sin(xy)= sin (2y) \: \dfrac{dy}{dx} + \dfrac{dy}{dx} - x \: sin(xy)\dfrac{dy}{dx} }$

$\implies \sf{ y \: sin(xy)= \{ 1 + sin (2y) - x \: sin(xy) \} \: \dfrac{dy}{dx} }$

$\implies \sf{ \dfrac{ y \: sin(xy)}{ 1 + sin (2y) - x \: sin(xy)}= \dfrac{dy}{dx} }$

$\implies \sf{ \dfrac{dy}{dx} = \dfrac{ y \: sin(xy)}{ 1 + sin (2y) - x \: sin(xy) } }$

Answer 2

Answer:

Hope itt helpss bro!!!!!!