If $y^{3} + y = x^{2}$ $y^3+y=x^2$ , what is $\frac{d y}{d x}$ $(dy)/(dx)$ ?

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Answer 1 · 2017-04-13T18:13:21

$\frac{d y}{d x} = \frac{2 x}{3 y^{2} + 1}$ $dy/dx=(2x)/(3y^2+1)$

$y^{3} + y = x^{2}$ $y^3+y=x^2$

By differentiating w.r.t. $x$ $x$ ,

$\Rightarrow 3 y^{2} \frac{d y}{d x} + \frac{d y}{d x} = 2 x$ $Rightarrow 3y^2 dy/dx+ dy/dx=2x$

By factoring out $\frac{d y}{d x}$ $dy/dx$ ,

$\Rightarrow (3 y^{2} + 1) \frac{d y}{d x} = 2 x$ $Rightarrow (3y^2+1) dy/dx=2x$

By dividing both sides by $(3 y^{2} + 1)$ $(3y^2+1)$ ,

$\Rightarrow \frac{d y}{d x} = \frac{2 x}{3 y^{2} + 1}$ $Rightarrow dy/dx=(2x)/(3y^2+1)$

I hope that this was clear.

If y^3+y=x^2y3+y=x2y^3+y=x^2, what is (dy)/(dx)dydx(dy)/(dx)?