Question

1. use implicit differentiation to find ( y ) for ( y^{3}+x^{3}=x^{3}y^{3}+1 ).

Explanation:

Step1: Differentiate both sides of the equation

Differentiate \(y^{3}+x^{3}=x^{3}y^{3}+1\) with respect to \(x\).
Using the sum rule \((u + v)^\prime=u^\prime+v^\prime\), the power rule \((x^n)^\prime = nx^{n - 1}\), and the product rule \((uv)^\prime=u^\prime v+uv^\prime\) (where \(u = x^{3}\) and \(v = y^{3}\)).
For the left - hand side: \(\frac{d}{dx}(y^{3}+x^{3})=\frac{d}{dx}(y^{3})+\frac{d}{dx}(x^{3})\). By the chain rule \(\frac{d}{dx}(y^{3}) = 3y^{2}y^\prime\) and \(\frac{d}{dx}(x^{3})=3x^{2}\).
For the right - hand side: \(\frac{d}{dx}(x^{3}y^{3}+1)=\frac{d}{dx}(x^{3}y^{3})+\frac{d}{dx}(1)\). Using the product rule, \(\frac{d}{dx}(x^{3}y^{3})=3x^{2}y^{3}+x^{3}\times3y^{2}y^\prime\) and \(\frac{d}{dx}(1) = 0\).
So we have \(3y^{2}y^\prime+3x^{2}=3x^{2}y^{3}+3x^{3}y^{2}y^\prime\).

Step2: Solve for \(y^\prime\)

First, move all terms with \(y^\prime\) to one side:
\(3y^{2}y^\prime-3x^{3}y^{2}y^\prime=3x^{2}y^{3}-3x^{2}\).
Factor out \(y^\prime\) on the left - hand side: \(y^\prime(3y^{2}-3x^{3}y^{2})=3x^{2}(y^{3}-1)\).
Then \(y^\prime=\frac{3x^{2}(y^{3}-1)}{3y^{2}(1 - x^{3})}\).
Simplify the fraction: \(y^\prime=\frac{x^{2}(y^{3}-1)}{y^{2}(1 - x^{3})}\).

Answer:

\(y^\prime=\frac{x^{2}(y^{3}-1)}{y^{2}(1 - x^{3})}\)