Which one of the following is false?

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(D) Let us evaluate each option:$A$: $\frac{d}{dx}[\sec^{-1}(\cosh x)] = \frac{1}{|\cosh x|\sqrt{\cosh^2 x - 1}} \cdot \sinh x = \frac{\sinh x}{\cosh

Vedclass · Accepted Answer

$\frac{d}{dx}[	an^{-1}(	an \frac{x}{2})] = \frac{1}{2}$

Vedclass · Answer

(D) Let us evaluate each option:$A$: $\frac{d}{dx}[\sec^{-1}(\cosh x)] = \frac{1}{|\cosh x|\sqrt{\cosh^2 x - 1}} \cdot \sinh x = \frac{\sinh x}{\cosh x \sinh x} = 	ext{sech } x$. (True)$B$: $\frac{d}{dx}[\cos^{-1}(	ext{sech } x)] = -\frac{1}{\sqrt{1-	ext{sech}^2 x}} \cdot (-	ext{sech } x 	anh x) = \frac{	ext{sech } x 	anh x}{	anh x} = 	ext{sech } x$. (True)$C$: $\frac{d}{dx}[	an^{-1}(\sinh x)] = \frac{1}{1+\sinh^2 x} \cdot \cosh x = \frac{\cosh x}{\cosh^2 x} = 	ext{sech } x$. (True)$D$: $\frac{d}{dx}[	an^{-1}(	an \frac{x}{2})] = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$. Since $\frac{1}{2} 
eq \sec x$,this statement is false.

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