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I have been given an exercise to compute the derivative of the following function:

$$y(x) = e^{-(y(x)+2)^2} \arctan(y(x)) $$

Now, what confuses me is that I don't know what's the variable by which we derive? Is this another way of writing:

$$y = e^{-(y+2)^2} \arctan(y), $$ or do I need to derive $y$ by $x$ somehow?

Thanks!

Madhav Nakar
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l0ner9
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  • Is this some sort of implicit function? When you write, for example, $\text{arctan}~{y}$, is that $y$ the same $y$ as on the left-hand side of the expression? – Matti P. Dec 10 '19 at 13:23
  • I'm not entirely sure, this is all I have been given. I suspect it may be an implicit function, and even if it is it's given in a strange way. – l0ner9 Dec 10 '19 at 13:25
  • It's a trick question. What the most appropriate way of attacking it is depends on the context in which this exercise was given. – Daniel Fischer Dec 10 '19 at 13:33
  • I don't know if it helps, but wolfram alpha says the only solution satisfying this given expression in Real set is $y=0$. – John Paul Dec 10 '19 at 13:36
  • More likely than not it is a trick question. The exercise states only the following:

    Compute the derivative of the following functions:

    After which follows a series of functions ranging from simple functions, parametric derivatives, logarithmic derivatives, and this one. That's why I suspect it might be an implicit function but it doesn't seem like one.

     – l0ner9 Dec 10 '19 at 13:38

2 Answers2

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The way it is written - it is an equation for $y$ which can be solved (in principle) to give you (one or more) fixed numbers. Thus $y$ is (one or more) constants and $y'=0$. If there was another term involving $x$ explicitly - perhaps you could have a non-zero derivatve.

am301
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By the product and the chain rule we get $$y'(x)=e^{-(y(x)+2)^2}(-2(y(x)+2))y'(x)\arctan(x))+e^{-(y(x)+2)^2}\frac{1}{1+y(x)^2}\times y'(x)$$ and solve this for $y'(x)$

Dr. Sonnhard Graubner
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