Question

use implicit differentiation to find \\( \frac { d y } { d x } \\).

\\( y ^ { 2 } + 5 x ^ { 3 } = 8 y - 2 x ^ { 2 } \\)

\\( \frac { d y } { d x } = \\)

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Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(y^{2}+5x^{3}\) and \(8y - 2x^{2}\) term - by - term.
For \(y^{2}\), use the chain rule: \(\frac{d}{dx}(y^{2})=2y\frac{dy}{dx}\).
For \(5x^{3}\), \(\frac{d}{dx}(5x^{3}) = 15x^{2}\).
For \(8y\), \(\frac{d}{dx}(8y)=8\frac{dy}{dx}\).
For \(-2x^{2}\), \(\frac{d}{dx}(-2x^{2})=-4x\).
So, \(2y\frac{dy}{dx}+15x^{2}=8\frac{dy}{dx}-4x\).

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

Move all terms with \(\frac{dy}{dx}\) to one side:
\(2y\frac{dy}{dx}-8\frac{dy}{dx}=-4x - 15x^{2}\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(2y - 8)=-4x - 15x^{2}\).
Then \(\frac{dy}{dx}=\frac{-4x - 15x^{2}}{2y - 8}=\frac{15x^{2}+4x}{8 - 2y}\).

Answer:

\(\frac{15x^{2}+4x}{8 - 2y}\)