0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Then how do you find $f(x)$ ? It's clearly needed for the following $\endgroup$ – Rebellos Jun 2 '16 at 17:07
  • $\begingroup$ It's not needed; do you remember implicit differentiation from Calc I? $\endgroup$ – GFauxPas Jun 2 '16 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 '16 at 17:33
  • $\begingroup$ @CharalamposFilippatos I take it that your question (from your comment) is answered? $\endgroup$ – Thomas Jun 2 '16 at 18:37
  • $\begingroup$ Yes for sure, also approved your answer a while ago :) Thanks a lot ! $\endgroup$ – Rebellos Jun 2 '16 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.