Question

find y'
( assuming y is the function of x )
sin (y) +3x^2 - y^3 = 4

thanks

Answer 1

cosy dy/dx +6x-3y^2 dy/de=0
dy/dx(cosy-3y^2)=-6x
dy/dx=-6x/cosy-3y^2
is your answer

Answer 2

Hey There!!!

Here, 
$\sin y+3x^2 - y^3 = 4 \\ \\ \textrm{Differentiating w.r.t. x} \\ \\ \implies \cos y \frac{dy}{dx} + 6x - 3y^2\frac{dy}{dx}=0 \\ \\ \implies (\cos y-3y^2)\frac{dy}{dx}= -6x \\ \\ \implies \boxed{{\frac{dy}{dx} = \frac{-6x}{\cos y - 3y^2}}$

Hope it helps
Purva
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