# Thread: What did I do wrong?

1. ## What did I do wrong?

I got this wrong on my test, my instructor did the work on the board and used a different approach than I did, I want to know what I did wrong in my approach that caused me to get the problem wrong.

Differentiate: $y = x^{e^{2x}}$

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My (flawed) approach:

$log_{x}y = e^{2x}$
differentiates to: $\frac{1}{y*ln(x)}y\prime = 2e^{2x}$
$y\prime=2e^{2x}*x^{e^{2x}}*ln(x)$
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$y\prime=x^{e^{2x}}(2e^{2x}ln(x)+\frac{e^{2x}}{x})$ and I can get it the instructor's way, but not my way, but I don't understand why my way failed me.

2. You have completely ignored the derivative of ln(x), treating it like a constant attached to 'y' on the left hand side.

3. Originally Posted by angel.white
I got this wrong on my test, my instructor did the work on the board and used a different approach than I did, I want to know what I did wrong in my approach that caused me to get the problem wrong.

Differentiate: $y = x^{e^{2x}}$

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My (flawed) approach:

${ \color{red}log_{x}y} = e^{2x}$
i do not like this approach, taking $\log_x$ is weird. but anyway, your problem is that you would need the product rule to differentiate what is in red, you did not use that. if you don't want to use the product rule, you must separate x and y

EDIT: TKHunny beat me to it

4. it seems so weird, but is it possible that
$\log_x y$
so we just need to apply chain, is that it?

to answer the original question, the easier way is to take the log or ln of both sides so that that
$e^{2x}$
will go to the side of log or ln, then you can apply implicit differentiation.. Ü

5. Originally Posted by kalagota
it seems so weird, but is it possible that
$\log_x y$
indeed it is weird. besides, if x is negative, we'd have a problem.

so we just need to apply chain, is that it?
we'd need to apply the product rule (implicitly).

to answer the original question, the easier way is to take the log or ln of both sides so that that
$e^{2x}$
will go to the side of log or ln, then you can apply implicit differentiation.. Ü
i agree