Question

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

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:

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Answer:

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