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exercises 3.8 implicit differentiation
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use implicit differentiation to determine \\( \frac{d y}{d x} \\) given the
equation \\( x^{6}+y^{4}=\cos y \\).
\\( \frac{d y}{d x}= \\)
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Explanation:

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

Differentiate \(x^{6}+y^{4}=\cos y\) term - by - term.
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), for the left - hand side: \(\frac{d}{dx}(x^{6})+\frac{d}{dx}(y^{4})\). We know that \(\frac{d}{dx}(x^{6}) = 6x^{5}\), and for \(\frac{d}{dx}(y^{4})\), by the chain rule \(\frac{d}{dx}(y^{4})=4y^{3}\frac{dy}{dx}\).
For the right - hand side, using the chain rule \(\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}\).
So we have \(6x^{5}+4y^{3}\frac{dy}{dx}=-\sin y\frac{dy}{dx}\).

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

First, move all terms with \(\frac{dy}{dx}\) to one side:
\(4y^{3}\frac{dy}{dx}+\sin y\frac{dy}{dx}=-6x^{5}\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(4y^{3}+\sin y)=-6x^{5}\).
Then, divide both sides by \((4y^{3}+\sin y)\) to get \(\frac{dy}{dx}=\frac{-6x^{5}}{4y^{3}+\sin y}\).

Answer:

\(\frac{dy}{dx}=\frac{-6x^{5}}{4y^{3}+\sin y}\)