Question

suppose that x and y are related by the given equation and use implicit differentiation to determine \\( \frac { d y } { d x } \\).
\\( x ^ { 4 } y + y ^ { 4 } x = 5 \\)
\\( \frac { d y } { d x } = \\)

Explanation:

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

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{4}\), \(v = y\) for the first - term \(x^{4}y\) and \(u = y^{4}\), \(v=x\) for the second - term \(y^{4}x\).
Differentiating \(x^{4}y\) gives \(4x^{3}y+x^{4}\frac{dy}{dx}\).
Differentiating \(y^{4}x\) gives \(4y^{3}\frac{dy}{dx}\cdot x + y^{4}\).
Differentiating the right - hand side (since \(\frac{d}{dx}(5)=0\)):
\(\frac{d}{dx}(x^{4}y + y^{4}x)=\frac{d}{dx}(5)\)
\(4x^{3}y+x^{4}\frac{dy}{dx}+4xy^{3}\frac{dy}{dx}+y^{4}=0\)

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

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

Answer:

\(\frac{-4x^{3}y - y^{4}}{x^{4}+4xy^{3}}\)