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Find the equation of the tangent line :

$$\ln{xy}= 2x $$ at point $( 1, e^2 )$

I end up with slope of $e^2$

so the equation will be $$ y= e^2(x-1) $$ $$ y = e^2x - e^2 $$

But the answer was just $$ y= e^2x $$ so appearantly they used $$ y=mx+b $$ instead of $$y=(x-x1)m +b $$

I know that my answer is wrong, but why?

Riccardo.Alestra
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user157908
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    Here in your last equation $b=e^2$ so it cancels. – coffeemath Jun 23 '14 at 13:37
  • @coffeemath to find be don't I have to substitute in the original equation? I forgot. – user157908 Jun 23 '14 at 13:40
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    The given point was $(1,e^2)$ so you can either say $a=1,b=e^2$ and use $y-b=m(x-a)$ or you can use $x_1=1,y_1=e^2$ and use the same formula replacing $a,b$. Suggestion: google on "point-slope formula". – coffeemath Jun 23 '14 at 13:43

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I think you mean: $y-y_1=m(x-x_1)$, so the equation becomes:

$$y-e^2=e^2(x-1)$$

$$y=e^2x-e^2+e^2$$

$$y=e^2x$$

rae306
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  • In the first equation you mention you only need one point, along with the slope $m$. This is IMO best suited for tangent line problems and then one can multiply out if one wants the other form. – coffeemath Jun 23 '14 at 13:46
  • Now it looks like a good explanation (+1) [I don't know if your edit might have already been under way when my last comment got typed...] – coffeemath Jun 23 '14 at 13:48
  • I think so. But thanks! ;-) – rae306 Jun 23 '14 at 13:49