problem with implicit derivative using ln

I have the following expression:

$$(xy)^{x^{2}}=(\tan y)^{xy^{3}}$$

With $$y$$ being an implicit and differentiable function of $$x$$. I want to find an expression for $$y'$$.

My first attempt is to use Ln function: $$x^{2}\ln(xy)=xy^{3}\ln(\tan y)$$.

But now I have two options: a) I use implicit differentiation (and other rules of differentiation) on the above equation. b) I rewrite the above expression as $$x\ln(xy)=y^{3}\ln(\tan y)$$, and then use implicit differentiation .

In my opinion, the two options should lead to the same final result. But for my surprise, this is not the case.

For a: $$y'=\dfrac{y^{3}\ln(\tan y)-x-2x\ln(xy)}{\dfrac{x^{2}}{y}-3xy^{2}\ln(\tan y)-xy^{3}\dfrac{\sec^{2}y}{\tan y}}$$

For b: $$y'=\dfrac{-1-2\ln(xy)}{\dfrac{x}{y}-3y^{2}\ln(\tan y)- y^{3}\dfrac{\sec^{2}y}{\tan y}}$$

What am I doing wrong? Which option is correct?

• They should lead to the same result. What did you get for each one?
– Ben
Jun 24 '21 at 22:12
• You asked "What am I doing wrong?" How are we supposed to know if you don't post what you did? Jun 24 '21 at 22:18
• Edit the comment with the results I have. Jun 24 '21 at 22:20
• Note that $y^3 \ln(\tan y)$, which appears in your first answer, is equal to $x \ln(xy)$. If you make that substitution and simplify, do your two answers turn out to be the same? Jun 25 '21 at 0:37

The part b is has a mistake, the $$2$$ shouldn't be there, it should be:
$$y'=\dfrac{-1-\ln(xy)}{\dfrac{x}{y}-3y^{2}\ln(\tan y)- y^{3}\dfrac{\sec^{2}y}{\tan y}}$$
After that, it's easy to check that both answer are equivalent. Just multiply by $$x$$ numerator and denominator of the RHS of b to get:
$$y'=\dfrac{-x-x\ln(xy)}{\dfrac{x^2}{y}-3x y^{2}\ln(\tan y)- x y^{3}\dfrac{\sec^{2}y}{\tan y}}$$
Now the denominators of a and b are the same, and the numerators are $$y^{3}\ln(\tan y)-x-2x\ln(xy)$$ and $$-x-x\ln(xy)$$ which are equal since we have $$y^{3}\ln(\tan y) = x\ln(xy)$$.