Question

Differentiate with respect to $x$
(a) $\tan h4x$
(b) $\sec h2x$

Answer 1

Formula:

1. $\sin hkx=\frac{e^{kx}-e^{-kx}}{2}$

2. $\cos hkx=\frac{e^{kx}+e^{-kx}}{2}$

3. $\frac{d}{dx}\frac{u}{v}=\frac{v\frac{d}{dx}u-u\frac{d}{dx}v}{v^2}$

(a) Given, $\tan h4x=\frac{\sin h4x}{\cos h4x}$

Therefore, $\tan h4x=\frac{e^{4x}-e^{-4x}}{e^{4x}+e^{-4x}}$

Differentiating $\tan h4x$ w.r.t $x$, we get

$\frac{d}{dx}\tan h4x=\frac{d}{dx}(\frac{e^{4x}-e^{-4x}}{e^{4x}-e^{-4x}})$

$=\frac{(e^{4x}+e^{-4x})\frac{d}{dx}(e^{4x}-e^{-4x})-(e^{4x}-e^{-4x})\frac{d}{dx}(e^{4x}+e^{-4x})}{(e^{4x}+e^{-4x}{)}^2}$

$=\frac{4(e^{4x}+e^{-4x}{)}^2-4(e^{4x}-e^{-4x}{)}^2}{(e^{4x}+e^{-4x}{)}^2}$

$=4[1-(\frac{e^{4x}-e^{-4x}}{e^{4x}+e^{-4x}}{)}^2]$

$=4(1-\tan h^{2}4x)$

(b) Given $\sec h2x=\frac{1}{\cos h2x}$

Therefore, $\sec h2x=\frac{2}{e^{2x}+e^{2x}}$

Differentiating $\sec h2x$w.r.t $x$, we get

$\frac{d}{dx}\sec h2x=\frac{d}{dx}(\frac{2}{e^{2x}+e^{-2x}})$

$=\frac{(e^{2x}+e^{-2x})\frac{d}{dx}2-2\frac{d}{dx}(e^{2x}+e^{-2x})}{(e^{2x}+e^{-2x}{)}^2}$

$=\frac{0-4(e^{2x}-e^{-2x})}{(e^{2x}+e^{-2x}{)}^2}$

$=-4\frac{1}{e^{2x}+e^{-2x}}.\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$

$=-4\sec h2x\tan h2x$