Question

find dy/dx of ax by^2=cosy​

Answer 1

GIVEN :–

$\\ \implies \tt (ax)(b {y}^{2} ) = \cos(y)$

TO FIND :–

• Value of dy/dx = ?

SOLUTION :–

$\\\implies \tt (ax)(b {y}^{2} ) = \cos(y)$

$\\\implies \tt (ab)(x{y}^{2} ) = \cos(y)$

• Now Differentiate with respect to 'x' –

$\\\implies \tt \dfrac{d \{(ab)(x{y}^{2} ) \}}{dx} = \dfrac{d \{\cos(y) \} }{dx}$

• Using identity –

$\\\longrightarrow \large{ \boxed{ \tt \dfrac{d \{kf(x) \}}{dx} =k \dfrac{d \{f(x) \} }{dx}}}$

• So that –

$\\\implies \tt (ab) \dfrac{d(x{y}^{2} )}{dx} = \dfrac{d \{\cos(y) \} }{dx}$

• Using identity –

$\\\longrightarrow \large{ \boxed{ \tt \dfrac{d(u.v)}{dx} =u\dfrac{dv}{dx} + v\dfrac{du}{dx}}}$

• And we know that –

$\\\longrightarrow \large{ \boxed{ \tt \dfrac{d( \cos x)}{dx} = - \sin(x) }}$

• So that –

$\\\implies \tt (ab) \bigg[ x\dfrac{d({y}^{2} )}{dx} + {y}^{2} \dfrac{dx}{dx}\bigg]= - \sin(y) \dfrac{d(y)}{dx}$

$\\\implies \tt (ab) \bigg[ x(2y)\dfrac{dy}{dx} + {y}^{2} \bigg]= - \sin(y) \dfrac{d(y)}{dx}$

$\\\implies \tt 2abxy \dfrac{dy}{dx} + ab {y}^{2} = - \sin(y) \dfrac{d(y)}{dx}$

$\\\implies \tt 2abxy \dfrac{dy}{dx} + ab {y}^{2} + \sin(y) \dfrac{d(y)}{dx} = 0$

$\\\implies \tt \dfrac{dy}{dx}(2abxy + \sin y) + ab {y}^{2} = 0$

$\\\implies \tt \dfrac{dy}{dx}(2abxy + \sin y) = - ab {y}^{2}$

$\\\implies \large { \boxed{\tt \dfrac{dy}{dx}= \dfrac{- ab {y}^{2}}{(2abxy + \sin y) } }}$

Answer 2

Correct Question :

• find dy/dx of ax + by^2=cosy

Given :

• ax + by^2=cosy

To Find :

• find the value of dy/dx

Solution :

We have ,

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

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