Difference between revisions of "Derivative of sech"
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Using the [[quotient rule]], the [[derivative of cosh]], and the definition of $\mathrm{tanh}$, we see | Using the [[quotient rule]], the [[derivative of cosh]], and the definition of $\mathrm{tanh}$, we see | ||
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
| − | \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z) &= \dfrac{-\sinh(z)}{\cosh(z)^2} \\ | + | \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z) &= \dfrac{0-\sinh(z)}{\cosh(z)^2} \\ |
&=-\mathrm{sech}(z)\mathrm{tanh}(z), | &=-\mathrm{sech}(z)\mathrm{tanh}(z), | ||
\end{array}$$ | \end{array}$$ | ||
Latest revision as of 11:47, 17 September 2016
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z)=-\mathrm{sech}(z)\mathrm{tanh}(z),$$ where $\mathrm{sech}$ denotes the hyperbolic secant and $\mathrm{tanh}$ denotes the hyperbolic tangent.
Proof
From the definition, $$\mathrm{sech}(z) = \dfrac{1}{\mathrm{cosh}(z)}.$$ Using the quotient rule, the derivative of cosh, and the definition of $\mathrm{tanh}$, we see $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z) &= \dfrac{0-\sinh(z)}{\cosh(z)^2} \\ &=-\mathrm{sech}(z)\mathrm{tanh}(z), \end{array}$$ as was to be shown.
