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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?

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    $\begingroup$ They should lead to the same result. What did you get for each one? $\endgroup$
    – Ben
    Jun 24 '21 at 22:12
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    $\begingroup$ You asked "What am I doing wrong?" How are we supposed to know if you don't post what you did? $\endgroup$
    – jjagmath
    Jun 24 '21 at 22:18
  • $\begingroup$ Edit the comment with the results I have. $\endgroup$
    – mmramame
    Jun 24 '21 at 22:20
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    $\begingroup$ 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? $\endgroup$ Jun 25 '21 at 0:37
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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 $y^{3}\ln(\tan y) = x\ln(xy)$.

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