Answer: The required value of [tex]\dfrac{dy}{dx}[/tex] is [tex]\dfrac{y}{1-y}.[/tex]
Step-by-step explanation: We are given to find the value of [tex]\dfrac{dy}{dx}[/tex] from the following equation :
[tex]\ln y=(\ln x)^2+2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Differentiating both sides of equation (i) with respect to x, we have
[tex]\dfrac{d}{dx}\ln y=\dfrac{d}{dx}((\ln x)^2+2)\\\\\\\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{d}{dx}(\ln x)^2+\dfrac{d}{dx}2\\\\\\\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=2\ln x\times\dfrac{d}{dx}\ln x+0\\\\\\\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=2\dfrac{\ln x}{x}\\\\\\\Rightarrow \dfrac{dy}{dx}=\dfrac{2y\ln x}{x}.[/tex]
Thus, the required value of [tex]\dfrac{dy}{dx}[/tex] in terms of x and y is [tex]\dfrac{2y\ln x}{x}.[/tex].
