Question

sin^{2} y + cos xy = k dy/dx ज्ञात कीजिए

Answer 1

Question :-

$\rm if \: \sf { \sin}^{2} y + \cos(xy) = k \: then \: find \: \frac{dy}{dx}$

Answer :-

$\to \boxed{\sf \frac{dy}{dx} = \frac{y \sin(xy)}{ \sin2y - x \sin(xy)} } \:$

Solution :-

we have,

$\mapsto \: \sf { \sin}^{2} y + \cos(xy) = k \: \: \\ \\ \rm \pink{ differentiate \: it \: with \: respect \: to \: x} \\ \\ \to \sf \frac{d}{dx} { \sin}^{2} y + \frac{d}{dx} \cos(xy) = \frac{d}{dx} k \\ \\ \boxed{ \because \sf \frac{d}{dx} (constant) = 0 \: and \: \frac{d}{dx} {x}^{n} = n {x}^{n - 1} } \\ \\ \boxed{ \sf\because \frac{d}{dx} \cos x = - \sin x } \\ \\ \to \sf 2 \sin y \frac{d}{dx} ( \sin y) - \sin(xy) \frac{d}{dx} xy = 0 \\ \\ \sf \: here \: we \: use \: chain \: rule \\ \sf that \: is \\ \boxed{ \to \sf \frac{d}{dx} xy = x \frac{d}{dx} y+ y \frac{d}{dx} x} \\ \\ \sf hence \\ \to \sf 2 \sin y \cos y \frac{dy}{dx} - \sin(xy) \{x \frac{dy}{dx} + y \} = 0 \\ \\ \boxed{\because \sf 2 \sin \theta \cos \theta = \sin2 \theta} \\ \\ \to \sf \sin2y \frac{dy}{dx} - x \sin(xy) \frac{dy}{dx} - y \sin(xy) = 0 \\ \\ \to \sf \frac{dy}{dx} \: \{ \sin2y - x \sin(xy) \} = y \sin (xy) \\ \\ \to \boxed{\sf \frac{dy}{dx} = \frac{y \sin(xy)}{ \sin2y - x \sin(xy)} }$

Answer 2

(dy/dx) =  ySinxy /(Sin2y - xSinxy)   यदि Sin²y   + Cosxy  = k

Step-by-step explanation:

dy/dx ज्ञात कीजिए

Sin²y   + Cosxy  = k

=> 2Siny (cosy) dy/dx   - Sinxy* (xdy/dx  + y)  = 0

=> (dy/dx)(2Sinycosy - xSinxy)  = ySinxy

=>  (dy/dx) =  ySinxy /(2Sinycosy - xSinxy)

=>  (dy/dx) =  ySinxy /(Sin2y - xSinxy)

और अधिक जानें :

sin(x²+5)"

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