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Show that $$\frac{d}{d x}\left(\cosh ^{-1} x\right)=\frac{1}{\sqrt{x^{2}-1}} \cdot \quad x>1$$...

Question

Show that $$\frac{d}{d x}\left(\cosh ^{-1} x\right)=\frac{1}{\sqrt{x^{2}-1}} \cdot \quad x>1$$

Show that $$\frac{d}{d x}\left(\cosh ^{-1} x\right)=\frac{1}{\sqrt{x^{2}-1}} \cdot \quad x>1$$



Answers

Prove the identity.
$ \cosh (-x) = \cosh x $ (This shows that $ cosh $ is an even function.)


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