First and Second derivatives differentiation

Question

The equation $x^5y + x +y^3 =3$ defines implicitly a function $y=g(x)$ near $x=1$. Compute $g(1)$ , $g'(1)$, and $g''(1)$.

If someone could show me the first few steps that would help.

Answer 1

you have the constraint $$x^5y + x +y^3 =3\tag 1$$

putting $x = 1,$ gives you $y+y^3 = 2$ and the only solution is $y = 1.$ differencing $(1)$ we get $$x^5\, \frac{dy}{dx} + 5x^4y+1+3y^2 \, \frac{dy }{dx}= 0 \tag 2$$ putting $x = 1, y = 1$ should give you $\frac{dy}{dx} = -\frac32.$

i will let you difference $(2)$ and sub the values $x = 1, y = 1, \frac{dy}{dx} = -3/2$ to get the value of $\frac{d^2y}{dx^2}.$