Question

1. find \\( \frac { d y } { d x } \\) by implicit differentiation \\( x ^ { 4 } + x ^ { 2 } y ^ { 2 } + y ^ { 3 } = 5 \\)

Explanation:

Step1: Differentiate each term

Differentiate \(x^{4}+x^{2}y^{2}+y^{3}\) with respect to \(x\).
For \(x^{4}\), using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), we have \(\frac{d}{dx}(x^{4}) = 4x^{3}\).
For \(x^{2}y^{2}\), use the product rule \((uv)^\prime=u^\prime v+uv^\prime\) where \(u = x^{2}\) and \(v=y^{2}\).
\(\frac{d}{dx}(x^{2}y^{2})=2xy^{2}+x^{2}\cdot2y\frac{dy}{dx}\).
For \(y^{3}\), using the chain rule \(\frac{d}{dx}(y^{n})=ny^{n - 1}\frac{dy}{dx}\), we get \(\frac{d}{dx}(y^{3})=3y^{2}\frac{dy}{dx}\).
The derivative of the constant \(5\) is \(0\). So, \(4x^{3}+2xy^{2}+2x^{2}y\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Group the terms with \(\frac{dy}{dx}\) together:
\((2x^{2}y + 3y^{2})\frac{dy}{dx}=-4x^{3}-2xy^{2}\).
Then \(\frac{dy}{dx}=\frac{-4x^{3}-2xy^{2}}{2x^{2}y + 3y^{2}}=\frac{-2x(2x^{2}+y^{2})}{y(2x^{2}+3y)}\).

Answer:

\(\frac{dy}{dx}=\frac{-2x(2x^{2}+y^{2})}{y(2x^{2}+3y)}\)