Question

Q) Find dy/dx 2x+3Y square=cosx ?​

Answer 1

we have to differentiate the following expression :-

$\rm 2x + 3 {y}^{2} = cos \: x$

$\rm 3 {y}^{2} = cos \: x - 2x$

$\rm {y}^{2} = \dfrac{cos \: x - 2x}{3} \\ \\ \rm {y} = \sqrt{\dfrac{cos \: x - 2x}{3}} ...(1)$

$\rm \longrightarrow 3 \dfrac{d( {y}^{2})}{dx} = \dfrac{d( cos \: x - 2x)}{dx}$

$\rm \longrightarrow 6y \dfrac{d{y}}{dx} = - sinx - 2 \\ \\\rm \longrightarrow 6y \dfrac{d{y}}{dx} = -( sinx + 2 ) \\ \\\rm \longrightarrow \dfrac{d{y}}{dx} = \dfrac{-( sinx + 2 )}{6y}$

You can put the value of y from equation (1).

Answer :-

$\rm\dfrac{d{y}}{dx} = \dfrac{-( sinx + 2 )}{6y}$

Answer 2

Answer:

$→2x + 3 {y}^{2} = \cos(x) \\ 2 + 6y \frac{dy}{dx} = -\sin(x) \\ 6y \frac{dy}{dx} = - \sin(x) - 2 \\ \boxed{ \frac{dy}{dx} = \frac{ - ( \sin(x) + 2) }{6y} }$

• dy/dx = -(sin(x)+2)/6y is the right answer.