Here is a table of the derivatives of some of the hyperbolic functions:
$f(x)$ |
$\displaystyle{\frac{df}{dx}}$ |
$\sinh(bx)$ |
$b\cosh(bx)$ |
$\cosh(bx)$ |
$b\sinh(bx)$ |
$\tanh(bx)$ |
$\simplify{b*sech(bx)^2}$ |
a) $\displaystyle{f(x)=\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}}$
Using the product rule we obtain:
\[\frac{df}{dx} = \simplify[std]{{n} * (x ^ {(n -1)}) * sinh({a1} * x + {b1}) + {a1} * (x ^ {n}) * cosh({a1} * x + {b1})}\]
b) $\displaystyle{f(x)=\tanh(\simplify[std]{{a}x+{b}})}$
Using the table we can immediately write the derivative as:
\[\frac{df}{dx} = \simplify[std]{{a}*sech({a}x+{b})^2}\]
c) $\displaystyle{f(x)=\ln(\cosh(\simplify[std]{{a2}x+{b2}})}$
Here we employ the chain rule. We set $u=\cosh(\simplify[std]{{a2}x+{b2}})$, such that $f(x)=f(u)=\ln u$. Then, according to the chain rule:
\[\frac{df}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \]
Evaluating the derivatives on the right-hand side: $\displaystyle{\frac{du}{dx}=\simplify[std]{{a2}*sinh({a2}x+{b2})}}$ and $\displaystyle{\frac{df}{du}=\frac{1}{u}=\frac{1}{\cosh(\simplify[std]{{a2}x+{b2}})}}$.
Then, inserting these into the chain rule gives:
\[\begin{eqnarray*}\frac{df}{dx} &=& \frac{\simplify[std]{{a2} * sinh({a2} * x + {b2})}}{\cosh(\simplify[std]{{a2}x+{b2}})} \\ &=& \simplify[std]{{a2} * tanh({a2} * x + {b2})}\end{eqnarray*}\]