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If it is given that $$ e^xf(x) = 2 + \int_0^x\sqrt{1+x^4}\,dx $$

then what is the value of $ \dfrac {d} {dx} \Big(f^{-1}(x)\Big)\Bigg|_{x=2} $

Where I am stuck:

Now, since we are to evaluate $ \dfrac {d} {dx} \Big(f^{-1}(x)\Big)\Bigg|_{x=2} $, all we need to evaluate is $ f'(2) $. Our answer will be the reciprocal of this.

So I differentiated the given equation:

$$ e^x(f(x) + f'(x)) = \sqrt{1+x^4}\ $$

But to find $f'(2)$ we also need to evaluate $f(2)$ which seems to be an insane thing to do. So what to do?

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We need $f^{-1}(2)$, which is $0$. – André Nicolas Aug 18 '13 at 5:58
Put $x=0$. The right-hand side is $2$. The left-hand side is $e^0f(0)$. So $f(0)=2$, and therefore $f^{-1}(2)=0$. – André Nicolas Aug 18 '13 at 6:02
I got that part. That was a dumb thing. So I deleted the comment right away. Still working on the main problem though - but please don't tell how we're gonna use $f^{-1}(2)$ right away. – Parth Thakkar Aug 18 '13 at 6:02
Ok I give up. Some more hint please! – Parth Thakkar Aug 18 '13 at 6:06
Don't worry! This is a homework-style question, I will not write out a solution, at least not for quite a while. I just wanted to deal with the thing that was stopping you. – André Nicolas Aug 18 '13 at 6:07
up vote 1 down vote accepted

Hint: We don't need $f(2)$, we need $f^{-1}(2)$, which is $0$.

Now go ahead and find $f'(0)$, like you (sort of) started to do.

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Ok. I got till here $f'(0) = f'(f^{-1}(2)) = -1$. Now I gotta use this to find $(f^{-1})'(2)$...but I don't understand how to go further. – Parth Thakkar Aug 18 '13 at 6:21
Using Fundamental Theorem of Calculus, you showed that $f'(0)=-1$. It follows that $\frac{d}{dx}f^{-1}(x)$ is $\frac{1}{-1}$ at $x=2$. So basically you had reached the end of the calculation. – André Nicolas Aug 18 '13 at 6:27
I just, just got it. I got stuck at $f'(f^{-1}(2))$, coz over there I didn't have any $x$ term explicitly. Then i rewrote and got it. Thanks for spending nearly half an hour for this 1 problem!!! – Parth Thakkar Aug 18 '13 at 6:29
You are welcome. Life would be easier if books used $f^{-1}(y)$. – André Nicolas Aug 18 '13 at 6:31

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