Question

Use the quotient rule, or otherwise, to prove that
$\frac{d}{dx}\left(\sec hx\right)=-\sec hx\tan hx$

Answer 1

We know that sechx$=\frac{1}{\cosh x}$
differentiate the given equation with respect to 'x' using quotient rule
$=\frac{\cosh x\left(f^{{}^{\prime }}\left(1\right)\right)+\left(1\right)\left(f^1\left(\cosh x\right)\right)}{{\cosh }^{2}x}$
$=\frac{\cosh x\left(0\right)-\sinh x}{{\cosh }^{2}x}$
$=\frac{0-\sinh x}{{\cosh }^{2}x}$
$=\frac{-\sinh x}{{\cosh }^{2}x}$
$=\frac{-\sinh x}{\cosh x\cosh x}$
$=-\tanh x\times \frac{1}{\cosh x}$
$=-\tanh x\times sechx$
$=-\tanh xsechx$
Therefore we have proved that
$\frac{d}{dx}\left(sechx\right)=-\tanh xsechx$