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My professor wrote this problem on the board as a challenge:

Find the tangent line at (0,0) to the curve defined implicitly below. $$\ln(1+x+y)=\left( x^{42} e^y + \cos(xy)\sin(xy)\right)^{2015} \left( ye^{x\cos y} + 429 \sin(y \cos(x))\right)^{257} + 2x$$

I tried differentiating the equation, but I gave up because it was just too messy. Is there an easier way?

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    $\begingroup$ i don't believe this exercise serves any useful purpose; in fact it is useless. you might want to find the tangent line of $x^2 - xy + y^2 = 1$ . $\endgroup$ – abel Mar 28 '15 at 12:39
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    $\begingroup$ My professor said that there is a trick to it. I am already comfortable with problems like yours. $\endgroup$ – TorsionSquid Mar 28 '15 at 12:40
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    $\begingroup$ your prof must be a genius; don't take his/her word for it. press for details. come back and tell us all about it. $\endgroup$ – abel Mar 28 '15 at 12:42
  • $\begingroup$ Ok. I won't be seeing him until next week. I was just very curious, since he said that he can do this in his head in half a minute! $\endgroup$ – TorsionSquid Mar 28 '15 at 12:43
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    $\begingroup$ WA $\endgroup$ – Arpan Mar 28 '15 at 12:44
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Write $$\ln(1+x+y)=f(x,y)^{2015}g(x,y)^{257}+2x$$

Now differentiate:

$$\frac{1+y'}{1+x+y}=2015y'f'(x,y)f(x,y)^{2014}g(x,y)^{257}+257y'g'(x,y)g(x,y)^{256}f(x,y)^{2015}+2$$

Since $f(0,0)=g(0,0)=0$, $$1+y'=2$$

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  • $\begingroup$ Great! I think we can just let $F=f^{2015}$ and $G=g^{257}$ and deal with $F$ and $G$, right? Then $F'G+FG'$ at $(0,0)$ is $0$. $\endgroup$ – TorsionSquid Mar 28 '15 at 12:52

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