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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?2012-08-22
  • 1
    http://en.wikipedia.org/wiki/Inverse_function#Inverses_and_composition2012-08-22
  • 1
    I got it!!! $\cosh^{-1}(\cosh x) = x$ sorry to bug you..2012-08-22

2 Answers 2

8

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.2012-08-22
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    @alok What is $f^{-1}(f(x))$ if $f(x)$ is invertable?2012-08-22
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    @alok: What is $\ln e^x$? More generally, if $f$ is any invertible function, what is $f^{-1}(f(x))$?2012-08-22
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    Composition of a function.Chain Rule isn't it? Not sure2012-08-22
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    @BrianM.Scott $ln e^x = x$.2012-08-22
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    @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 $x$.2012-08-22
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    @alok Also, if we let $f(x)=\ln x$ and $f^{-1}(x)= e^x$ we also find that $f^{-1}(f(x))=\ln e^x =x$ as well.2012-08-22
6

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 $2|x|$.