Difference between revisions of "Derivative of Gudermannian"
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| Line 6: | Line 6: | ||
==Proof== | ==Proof== | ||
From the definition, | From the definition, | ||
| − | $$\mathrm{gd}(x) = \displaystyle\int_0^x \ | + | $$\mathrm{gd}(x) = \displaystyle\int_0^x \mathrm{sech}(t) \mathrm{d}t.$$ |
| − | + | Using the [[fundamental theorem of calculus]], | |
$$\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{gd}(x) = \dfrac{1}{\cosh x} = \mathrm{sech}(x),$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{gd}(x) = \dfrac{1}{\cosh x} = \mathrm{sech}(x),$$ | ||
as was to be shown. | as was to be shown. | ||
Revision as of 22:08, 19 September 2016
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{gd}(x)=\mathrm{sech}(x),$$ where $\mathrm{gd}$ denotes the Gudermannian and $\mathrm{sech}$ denotes the hyperbolic secant.
Proof
From the definition, $$\mathrm{gd}(x) = \displaystyle\int_0^x \mathrm{sech}(t) \mathrm{d}t.$$ Using the fundamental theorem of calculus, $$\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{gd}(x) = \dfrac{1}{\cosh x} = \mathrm{sech}(x),$$ as was to be shown.
