Derivatives and Logarithms 103

Given that $\dfrac {x^2}{y^2} = x \ln y$ ,

$\begin{aligned} \implies x & = y^2 \ln y = \frac {y^2}2 \ln y^2 & \small \blue{\text{Let }u = \ln y^2} \\ \implies u e^u & = 2x \\ W(ue^u) & = W(2x) & \small \blue{\text{Since }W(ae^a) = a} \\ \implies u & = W(2x) \\ \ln y^2 & = W(2x) \\ \implies y & = \sqrt{e^{W(2x)}} \end{aligned}$

From $x = y^2 \ln y$ , differentiate both sides with respect to $x$ :

$\begin{aligned} 1 & = \left(2y \ln y + \frac {y^2}y \right) \frac {dy}{dx} \\ \implies \frac {dy}{dx} & = \frac 1{(\blue{\ln y^2} + 1)y} & \small \blue{\text{Note that }\ln y^2 = W(2x)} \\ & = \frac 1{(W(2x)+1)\sqrt{e^{W(2x)}}} \end{aligned}$

Therefore $a = 2$ and $a^2 = \boxed 4$ .

Note: This can be done by rearranging the starting equation first, but I did it without that.

First, we can differentiate both sides using implicit differentiation: $\frac{d}{dx}\left(\frac{x^2}{y^2}\right)=\frac{d}{dx}(x\ln(y)) \Longrightarrow \frac{2x}{y^2} - \frac{2x^2}{y^3}\times \frac{dy}{dx} = \ln(y) + \frac{x}{y}\times \frac{dy}{dx}$

Then, we can rearrange to find $\Large \frac{dy}{dx}$ : $\frac{2x}{y^2} - \frac{2x^2}{y^3}\times \frac{dy}{dx} = \ln(y) + \frac{x}{y}\times \frac{dy}{dx} \Longrightarrow \frac{2x}{y^2} - \ln(y) = \frac{dy}{dx}\left( \frac{2x^2}{y^3}+ \frac{x}{y}\right)$ $\Longrightarrow \boxed{\frac{dy}{dx} = \frac{\frac{2x}{y^2} - \ln(y)}{\frac{2x^2}{y^3}+ \frac{x}{y}} = \frac{2xy-y^3\ln(y)}{2x^2+xy^2}}$

However, the requirement is to express $\Large \frac{dy}{dx}$ in terms of $x$ , not in terms of $y$ and $x$ . This means that we need to find out what $y$ is in terms of $x$ . We can do this by rearranging the starting equation: $\frac{x^2}{y^2} = x\ln(y) \Longrightarrow x = y^2\ln(y) \Longrightarrow x = \ln(y)e^{2\ln(y)}$

We can almost use the product log now! Once we multiply by $2$ to make the coefficient and exponent of $e$ the same, we can use it: $x = \ln(y)e^{2\ln(y)} \Longrightarrow 2x = \blue{2\ln(y)}e^{\blue{2\ln(y)}} \Longrightarrow W(2x) = \blue{2\ln(y)} \Longrightarrow \boxed{y = e^{\frac{W(2x)}{2}} = \sqrt{\frac{2x}{W(2x)}}}$

Finally, we have an expression for $y$ ! Now, we can plug this into our expression for $\Large \frac{dy}{dx}$ : $\begin{aligned} \frac{2xy-y^3\ln(y)}{2x^2+xy^2} \Longrightarrow \frac{2x\sqrt{\frac{2x}{W(2x)}}-\sqrt{\frac{2x}{W(2x)}}^3\ln(\sqrt{\frac{2x}{W(2x)}})}{2x^2+x\sqrt{\frac{2x}{W(2x)}}^2} &= \frac{2x\sqrt{\frac{2x}{W(2x)}}-\frac{2x}{W(2x)}\sqrt{\frac{2x}{W(2x)}}\ln(\frac{2x}{W(2x)})}{2x^2+\frac{2x^2}{W(2x)}} \\ &= \frac{2\sqrt{\frac{2x}{W(2x)}}-\frac{2}{W(2x)}\sqrt{\frac{2x}{W(2x)}}W(2x)}{2x+\frac{2x}{W(2x)}} \\ &= \frac{\sqrt{\frac{2x}{W(2x)}}}{\frac{2xW(2x)+2x}{W(2x)}} \\ &= \frac{W(2x)\sqrt{\frac{2x}{W(2x)}}}{2x(W(2x)+1)} \\ &= \sqrt{\frac{2x}{W(2x)}}\times \frac{W(2x)}{2x}\times \frac{1}{W(2x)+1} \\ \end{aligned}$

This may not be looking much like our needed expression, but look closely and you can see how to alter it:

$\sqrt{\frac{2x}{W(2x)}}\times \frac{W(2x)}{2x} = e^{\frac{W(2x)}{2}}\times e^{-W(2x)} = e^{\frac{-W(2x)}{2}} = \frac{1}{\sqrt{e^{W(2x)}}}$

Much better! Plugging this into our expression and we can satisfy the expression above: $\sqrt{\frac{2x}{W(2x)}}\times \frac{W(2x)}{2x}\times \frac{1}{(W(2x)+1)} \Longrightarrow \boxed{\frac{1}{(W(\blue{2}x)+1)\sqrt{e^{W(\blue{2}x)}}}}$

Perfect. With the expression satisfied, we can see that $a = 2$ and the final answer is $a^2 = 2^2 = \boxed{4}$