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?
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 $y^{3}\ln(\tan y) = x\ln(xy)$.