Question

ax + by^{2} = cosy dy/dx ज्ञात कीजिए

Answer 1

Question :-

$\to \sf if \: ax + b {y}^{2} = \cos y \: find\: \frac{dy}{dx}$

Answer :-

$\boxed{ \to \sf \frac{dy}{dx} = \frac{ - a}{2by + \sin y} } \:$

Solution :-

We have ,

$\to \sf \: ax + b {y}^{2} = \cos y \\ \\ \sf \red{differentiate \: }\green{ with} \blue{ \: respect }\pink{ \: to \: x} \: \\ \\ \to \sf \frac{d}{dx} ax + \frac{d}{dx} b {y}^{2} = \frac{d}{dx} \cos y \\ \\ \: \: \: \boxed{ \because \sf \frac{d}{dx} {x}^{n} = n {x}^{n - 1} \: and \: \frac{d}{dx} \cos x = - \sin x} \\ \\ \to \sf \: a \: + 2by \frac{dy}{dx} = - \sin y \frac{dy}{dx} \\ \\ \to \sf 2by \frac{dy}{dx} + \sin y \frac{dy}{dx} = - a \\ \\ \to \sf \frac{dy}{dx} \{2by + \sin y \} = - a \\ \\ \boxed{ \to \sf \frac{dy}{dx} = \frac{ - a}{2by + \sin y} }$

Answer 2

dy/dx = -a/(2by + Siny)  ax + by² = cosy

Step-by-step explanation:

dy/dx ज्ञात कीजिए

ax + by² = Cosy

a + b(2y) dy/dx  = - Siny (dy/dx)

=> a + 2by(dy/dx)  + Siny(dy/dx) = 0

=> (dy/dx)(2by + siny) = -a

=> dy/dx = - a/(2by + Siny)

और अधिक जानें :

sin(x²+5)"

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sin (ax+b) फलन का अवकलन कीजिए

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