Recall the relationship of the hyperbolic trig functions to the standard trig functions. I have a series on trigonometry and a post which explains these definitions.
\[\begin{align}
&\sinh x = -i \sin (i x) &\qquad\qquad& \sech \ x = \sec (i x) \\[10pt]
&\cosh x = \cos (i x) &\qquad\qquad& \csch \ x = i \csc (i x) \\[10pt]
&\tanh x = -i \tan (i x) &\qquad\qquad& \coth x = i \cot (i x)
\end{align}\]
Hyperbolic Sine
\[\begin{align}
\frac{d}{dx} \sinh (x)
&= \frac{d}{dx} \left [ - i \sin (i x) \right ] \\[10pt]
&= -i \cdot i \cos (i x) \\[10pt]
&= \cos (i x) \\[10pt]
&= \cosh(x)
\end{align}\]
Hyperbolic Cosine
\[\begin{align}
\frac{d}{dx} \cosh (x)
&= \frac{d}{dx} \left [ \cos (i x) \right ] \\[10pt]
&= - i \sin (i x) \\[10pt]
&= \sinh(x)
\end{align}\]
Hyperbolic Tangent
\[\begin{align}
\frac{d}{dx} \tanh (x)
&= \frac{d}{dx} \left [ - i \tan (i x) \right ] \\[10pt]
&= -i \cdot i \sec^2 (i x) \\[10pt]
&= \sec^2 (i x) \\[10pt]
&= \sech^2 (x)
\end{align}\]
Hyperbolic Secant
\[\begin{align}
\frac{d}{dx} \sech (x)
&= \frac{d}{dx} \left [ \sec (i x) \right ] \\[10pt]
&= i \tan (i x) \sec (i x) \\[10pt]
&= - \tanh(x) \sech(x)
\end{align}\]
Hyperbolic Cosecant
\[\begin{align}
\frac{d}{dx} \csch (x)
&= \frac{d}{dx} \left [ - i \csc (i x) \right ] \\[10pt]
&= -i \cdot i \cot (i x) \csc(i x) \\[10pt]
&= - i \cot (i x) \cdot i \csc(i x) \\[10pt]
&= - \coth(x) \csch(x)
\end{align}\]
Hyperbolic Cotangent
\[\begin{align}
\frac{d}{dx} \coth (x)
&= \frac{d}{dx} \left [ i \cot (i x) \right ] \\[10pt]
&= - i \cdot i \csc^2 (i x) \\[10pt]
&= - \csc^2 (i x) \\[10pt]
&= - \csch^2 (x)
\end{align}\]