Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 24

Show that the formulas for the derivatives of the functions a.) $\cos hx$, b.) $\tan hx$, c.) $\csc hx$ ,d.) $\sec hx$ and e.) $\cot hx$

a.) $\cos hx$


$
\begin{equation}
\begin{aligned}

\cos hx =& \frac{e^x + e^{-x}}{2}
\\
\\
\frac{d}{dx} (\cos hx) =& \frac{d}{dx} \left( \frac{e^x + e^{-x}}{2} \right)
\\
\\
\frac{d}{dx} (\cos hx) =& \frac{\displaystyle (2) \frac{d}{dx} (e^x + e^{-x}) - (e^x + e^{-x}) \frac{d}{dx} (2) }{(2)^2}
\\
\\
\frac{d}{dx} (\cos hx) =& \frac{2 [e^x + (-e^{-x})]}{4}
\\
\\
\frac{d}{dx} (\cos hx) =& \frac{e^x - e^{-x}}{2}

\end{aligned}
\end{equation}
$


We know that $\displaystyle \sin h(x) = \frac{e^x - e^{-x}}{2}$, so

$\displaystyle \frac{d}{dx} (\cos hx) = \sin hx$

b.) $\tan hx$


$
\begin{equation}
\begin{aligned}

\tan hx =& \frac{\sin hx}{\cos hx}
\\
\\
\frac{d}{dx} \tan hx =& \frac{d}{dx} \left( \frac{\sin hx}{\cos hx} \right)
\\
\\
\frac{d}{dx} \tan hx =& \frac{\displaystyle (\cos hx) \frac{d}{dx} (\sin hx) - (\sin hx) \frac{d}{dx} (\cos hx) }{(\cos hx)^2}
\\
\\
\frac{d}{dx} \tan hx =& \frac{(\cos hx) (\cos hx) - (\sin hx)(\sin hx)}{\cos h^2 x}
\\
\\
\frac{d}{dx} \tan hx =& \frac{\cos h^2 x - \sin h^2 x}{\cos h^2 x}


\end{aligned}
\end{equation}
$



We know that $\cos h^2 x - \sin h^2 x = 1$, so


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} \tan hx =& \frac{1}{\cos h^2 x}
\\
\\
\frac{d}{dx} \tan hx =& \sec h^2 x

\end{aligned}
\end{equation}
$


c.) $\csc hx$


$
\begin{equation}
\begin{aligned}

\csc hx =& \frac{1}{\sin hx}
\\
\\
\csc hx =& (\sin hx)^{-1}
\\
\\
\frac{d}{dx} (\csc hx) =& \frac{d}{dx} (\sin hx)^{-1}
\\
\\
\frac{d}{dx} (\csc hx) =& -(\sin hx)^{-2} \frac{d}{dx} (\sin hx)
\\
\\
\frac{d}{dx} (\csc hx) =& - (\sin hx)^{-2} (\cos hx)
\\
\\
\frac{d}{dx} (\csc hx) =& \frac{-1}{\sin h^2 x} \cdot \cos hx
\\
\\
\frac{d}{dx} (\csc hx) =& \frac{- \cos hx}{\sin hx} \cdot \frac{1}{\sin hx}
\\
\\
\frac{d}{dx} (\csc hx) =& - \cot hx \csc hx


\end{aligned}
\end{equation}
$


d.) $\sec hx$


$
\begin{equation}
\begin{aligned}

\sec hx =& \frac{1}{\cos hx}
\\
\\
\sec hx =& (\cos hx)^{-1}
\\
\\
\frac{d}{dx} (\sec hx) =& \frac{d}{dx} (\cos hx)^{-1}
\\
\\
\frac{d}{dx} (\sec hx) =& - (\cos hx)^{-2} \frac{d}{dx} (\cos hx)
\\
\\
\frac{d}{dx} (\sec hx) =& - (\cos hx)^{-2} (\sin hx)
\\
\\
\frac{d}{dx} (\sec hx) =& \frac{-1}{\cos h^2 x} \cdot \sin hx
\\
\\
\frac{d}{dx} (\sec hx) =& \frac{- \sin hx}{\cos hx} \cdot \frac{1}{\cos hx}
\\
\\
\frac{d}{dx} (\sec hx) =& - \tan hx \sec hx

\end{aligned}
\end{equation}
$


e.) $\cot hx$


$
\begin{equation}
\begin{aligned}

\cot hx =& \frac{\csc hx}{\sin hx}
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{d}{dx} \left( \frac{\cos hx}{\sin hx} \right)
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{\displaystyle (\sin hx) \frac{d}{dx} (\cos hx) - (\cos hx) \frac{d}{dx} (\sin hx)}{(\sin hx)^2}
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{(\sin hx)(\sin hx) - (\cos hx)(\cos hx)}{\sin h^2x}
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{\sin h^2 x - \cos h^2 x}{\sin h^2 x}
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{- (\cos h^2 x - \sin h^2 x)}{\sin h^2 x}

\end{aligned}
\end{equation}
$


We know that $\cos h^2 x - \sin h^2 x = 1$


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (\cot hx) =& \frac{-1}{\sin h^2 x}
\\
\\
\frac{d}{dx} (\cot hx) =& - \csc h^2 x

\end{aligned}
\end{equation}
$