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I've been sitting on this problem for a while, hopefully you guys could give me a lead on what the hell is going on :)

Let $f(x) = e^x + x$

Find the tangent line to $f^{-1}(y)$ (the inverse function) at $y=2$

Now I know that the derivative of the inverse at some $x$ is equal to the reciprocal of the derivative of the original function with the inverse function value at $x$ (hard explaining it, proves I don't understand it will enough huh? :) )

So I thought that finding what $f(x)=2$ is equal to and then move on from there...But I can't quite solve it either.

Any tips?? Thanks~

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  • $\begingroup$ It doesn't have an elementary inverse. Do you know what the product log function is? $\endgroup$ – recursive recursion Mar 27 '14 at 22:12
  • $\begingroup$ Nope...Perhaps my direction is wrong because the equation I wanted to solve in unsolvable according to my current 'course knowledge' ... Is there something else to do tho? $\endgroup$ – new one Mar 27 '14 at 22:14
  • $\begingroup$ I don't think you can solve it directly by taking the inverse, taking the derivative of the inverse, then plugging in $2$. I'm trying to think of a smarter way to do it... $\endgroup$ – recursive recursion Mar 27 '14 at 22:16
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I know that the derivative of the inverse at some x is equal to the reciprocal of the derivative of the original function with the inverse function value at x.

$(e^x)'=e^x\iff\ln'x=\dfrac1{e^{\ln x}}=\dfrac1x \\ \sin'x=+\cos x=+\sqrt{1-\sin^2x}\iff\arcsin'x=+\dfrac1{\sqrt{1-\sin^2(\arcsin x)}}=+\dfrac1{\sqrt{1-x^2}} \\ \cos'x=-\sin x=-\sqrt{1-\cos^2x}\iff\arccos'x=-\dfrac1{\sqrt{1-\cos^2(\arccos x)}}=-\dfrac1{\sqrt{1-x^2}} \\ \tan'x=1+\tan^2x\iff\arctan'x=\dfrac1{1+\tan^2(\arctan x)}=\dfrac1{1+x^2}$


hard explaining it

$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{f(x)-f(a)}{f^{-1}\Big(f(x)\Big)-f^{-1}\Big(f(a)\Big)}=\lim_{x\to a}\frac1{\dfrac{f^{-1}\Big(f(x)\Big)-f^{-1}\Big(f(a)\Big)}{f(x)-f(a)}}$$

$$=\frac1{\lim_{x\to a}\dfrac{f^{-1}\Big(f(x)\Big)-f^{-1}\Big(f(a)\Big)}{f(x)-f(a)}}=\frac1{\Big[f^{-1}\Big]'\Big(f(a)\Big)}$$


Let $f(x) = e^x + x$. So I thought that finding what $f(2)$ is equal to.

$f(2)=e^2+2$. — I assume you rather meant $\Big[f^{-1}\Big]'\Big(f(2)\Big)$. In which case, the answer is $\dfrac1{f'(2)}$ $=\dfrac1{\big(e^x+x\big)'_{x=2}}=\dfrac1{\big(e^x+1\big)_{x=2}}=\dfrac1{e^2+1}$

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  • $\begingroup$ Thanks a lot for the explanations, they are very clear ( wish that the lecturer would show these examples ) Anyway, Wolframalpha gave this result : 2 - W[E^2] - 2 (1 - E^2 W'[E^2]) + x (1 - E^2 W'[E^2]) $\endgroup$ – new one Mar 28 '14 at 8:20
  • $\begingroup$ I'm guessing that is the value of $f^{-1}\big(2\big)$ or $\Big[f^{-1}\Big]'\big(2\big)$, expressed in terms of the Lambert W function. $\endgroup$ – Lucian Mar 28 '14 at 8:46
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I think you are on the right track but may need a bit of complex analysis help as "recursive recursion" suggests.

If you use the relation $$ \tfrac{dy}{dx} = \frac{1}{\frac{dx}{dy}} $$ where $y = f(x)$, you get $$ m := \frac{dx}{dy} = \frac{1}{e^x+1} = \frac{1}{y-x+1} $$

There is no straightforward relation for $x$ in terms of $y$ but you can try the complex analysis solution (see Lambert W-Function) $$ x = y - W(e^y) $$ If we consider principal solution only, at $y=2$ we have $W(e^2)=1.557$ and $x = 0.443$. So the slope of the tangent is $m = 0.391$ (if my estimates are correct).

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