Difference between revisions of "Derivative of sech"
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| − | + | ==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 [[sech|hyperbolic secant]] and $\mathrm{tanh}$ denotes the [[tanh|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. | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
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.
