find \\( \\frac{d y}{d x} \\) in terms of \\( x \\) and \\( y \\) if \\( x \\ln y+y^{6}=4 \\ln x \\).\n\\(…

Explanation:

Step1: Differentiate both sides

Differentiate (x\ln y + y^{6}=4\ln x) with respect to (x) using product rule ((uv)^\prime = u^\prime v+uv^\prime) (for (x\ln y) where (u = x), (v=\ln y)), chain rule ((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)) (for (\ln y) and (y^{6})) and ((\ln x)^\prime=\frac{1}{x}).

Left - hand side: (\frac{d}{dx}(x\ln y)+\frac{d}{dx}(y^{6})=\ln y + x\cdot\frac{1}{y}\cdot\frac{dy}{dx}+6y^{5}\cdot\frac{dy}{dx})

Right - hand side: (\frac{d}{dx}(4\ln x)=\frac{4}{x})

So, (\ln y+\frac{x}{y}\frac{dy}{dx}+6y^{5}\frac{dy}{dx}=\frac{4}{x})

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

Group the terms with (\frac{dy}{dx}):

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

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

Multiply numerator and denominator by (xy) to get a common denominator:

[ \begin{align*} \frac{dy}{dx}&=\frac{4y - x y\ln y}{x^{2}+6xy^{6}} \end{align*} ]

Answer:

(\frac{4y - x y\ln y}{x^{2}+6xy^{6}})