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I'm revising things for the end-term semester exams, and Implicit Function Theorem is something that we studied for 1 month (theoretically), with every possible alternation on the theorem and it's theoretical existence, but we solved literally ZERO exercises, so I am having a big problem applying it. My question is the following :

Prove that the equation $e^{\cos xy} + x^2 + y - e = 0$ defines an unique implicit function $y=f(x)$ in an area of the point $(0,0)$ and then calculate $f'(0)$. After that, find the equation of the tangent of $e^{\cos xy} + x^2 + y - e = 0$ at $(0,0)$.

Now, my question is regarding the first part, how do I prove and find the unique defined implicit function $y=f(x)$ ? The rest can easily be done afterwards, since the calculation of the derivative is something elementary and finding the tangent :

At a $(x_0,y_0)$ of a curve $f(x,y) = 0$ the tangent is given by : $\langle \nabla f(x_0,y_0),(x-x_0,y-y_0)\rangle =0$ where $\langle a,b\rangle$ is the normal inner product in $\mathbb R^2$.

I would really appreciate any help-tip-walkthrough for how to prove the unique defined implicit function and how to find it.

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Let $h(x,y) = e^{\cos(xy)} + x^2 + y -e$. By calculation, if you insert $x=y=0$, you get that the point $(0,0)$ satisfies the equation $h(x,y)=0$ in question. To apply the implicit function theorem you, apart from this, have to show that, in that point, $\frac{\partial h}{\partial y} \neq 0$

But by the chain rule $$\frac{\partial h}{\partial y} =-e^{\cos(xy)}\sin(yx) x + 1 $$ which equals $1$ in $(0,0)$. So the assumptions of the theorem are fulfilled and the theorem implies that in a neighbourhood of $(0,0)$ the equation $h(x,y)=0$ has a unique differentiable solution $y = g(x)$.

(The theorem usually does not provide a closed form of the solution, nor does this necessarily exist).

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  • $\begingroup$ Then how do you find $f(x)$ ? It's clearly needed for the following $\endgroup$
    – Rebellos
    Jun 2, 2016 at 17:07
  • $\begingroup$ It's not needed; do you remember implicit differentiation from Calc I? $\endgroup$
    – GFauxPas
    Jun 2, 2016 at 17:20
  • $\begingroup$ Oh yes, the calculation of the implicit derivative is given by a fraction of the initial function. $\endgroup$
    – Rebellos
    Jun 2, 2016 at 17:33
  • $\begingroup$ @CharalamposFilippatos I take it that your question (from your comment) is answered? $\endgroup$
    – Thomas
    Jun 2, 2016 at 18:37
  • $\begingroup$ Yes for sure, also approved your answer a while ago :) Thanks a lot ! $\endgroup$
    – Rebellos
    Jun 2, 2016 at 18:38

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