Answer: The derivative would be [tex]\dfrac{dy}{dx}=\dfrac{(-2-4x)y}{x(4xy+1)}[/tex]
Step-by-step explanation:
Since we have given that
[tex]4xy + ln(x2y) = 7 [/tex]
We need to find the derivative implicity [tex]\dfrac{dy}{dx}[/tex]
so, we will assume y as constant when we derivative the function w.r.t x:
[tex]4xy+\ln x^2+\ln y=7\\\\4+4x\dfrac{dy}{dx}+\dfrac{1}{x^2}\times 2x+\dfrac{1}{y}\times \dfrac{dy}{dx}=0\\4+4x\dfrac{dy}{dx}+\dfrac{2}{x}+\dfrac{1}{y}\times \dfrac{dy}{dx}=0\\\\(4x+\dfrac{1}{y})\dfrac{dy}{dx}=-\dfrac{2}{x}-4\\\\\dfrac{4xy+1}{y}\dfrac{dy}{dx}=\dfrac{-2-4x}{x}\\\\\dfrac{dy}{dx}=\dfrac{(-2-4x)y}{x(4xy+1)}[/tex]
Hence, the derivative would be
[tex]\dfrac{dy}{dx}=\dfrac{(-2-4x)y}{x(4xy+1)}[/tex]
