Difference between revisions of "Derivative of Gudermannian"
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$$\mathrm{gd}(x) = \displaystyle\int_0^x \mathrm{sech}(t) \mathrm{d}t.$$ | $$\mathrm{gd}(x) = \displaystyle\int_0^x \mathrm{sech}(t) \mathrm{d}t.$$ | ||
Using the [[fundamental theorem of calculus]], | Using the [[fundamental theorem of calculus]], | ||
− | $$\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{gd}(x) | + | $$\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{gd}(x) = \mathrm{sech}(x),$$ |
as was to be shown. | as was to be shown. | ||
Latest revision as of 22:08, 19 September 2016
Theorem
The following formula holds: $$\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) = \mathrm{sech}(x),$$ as was to be shown.