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I'am suppose to show that $$\frac{\mathrm{d} }{\mathrm{d} x}[x \operatorname{cosh}^{-1}(\cosh x)] = 2x$$

And this is what i've tried.Upon differentiating the above function wrt $x$ using the product rule and applying the formula $\cosh^{-1}f(x) = \frac{f\prime(x)}{\sqrt{[f(x)]^2 - 1}}$, I end up getting $$x\frac{\sinh x}{\sqrt{(\cosh )^2 - 1}} + \cosh^{-1}(\cosh x)$$

How to proceed from here?

Thank You

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What is the capital of the country whose capital is Ottawa? – André Nicolas Aug 22 '12 at 15:51
I got it!!! $\cosh^{-1}(\cosh x) = x$ sorry to bug you.. – alok Aug 22 '12 at 16:00
up vote 8 down vote accepted

You’re working way too hard. Simplify! What is $\cosh^{-1}\big(\cosh x\big)$?

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Exactly.I've used $\cosh^{-1}f(x) = \frac{f\prime(x)}{\sqrt{[f(x)]^2 - 1}}$ and now i'am unable to simplify. – alok Aug 22 '12 at 15:51
@alok What is $f^{-1}(f(x))$ if $f(x)$ is invertable? – Argon Aug 22 '12 at 15:51
@alok: What is $\ln e^x$? More generally, if $f$ is any invertible function, what is $f^{-1}(f(x))$? – Brian M. Scott Aug 22 '12 at 15:53
Composition of a function.Chain Rule isn't it? Not sure – alok Aug 22 '12 at 15:53
@alok: Stop thinking about differentiation. That will come after you’ve simplified $\cosh^{-1}(\cosh x)$. Yes, $\ln e^x=x$; that’s because if $f(x)=e^x$, $\ln x=f^{-1}(x)$. The definition of $f^{-1}$ says that $f^{-1}(f(x))$ is always just $x$. – Brian M. Scott Aug 22 '12 at 15:55

If we are being fussy, the answer is not $2x$.

The function $\cosh x$ is an even function, like $\cos x$. Note that $\cosh^{-1} w$ is the non-negative number whose $\cosh$ is $w$. Now let $x$ be negative. Then $\cosh^{-1}(\cosh x)=-x$. Alternately, $\cosh^{-1}(\cosh x)=|x|$.

So the function we are trying to differentiate is in fact $x^2$ when $x\ge 0$ and $-x^2$ when $x\lt 0$. Its derivative is $2x$ when $x \ge 0$ and $-2x$ when $x\lt 0$, or alternately $2|x|$.

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