Difference between revisions of "Derivative of tanh"

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==Theorem==
<strong>[[Derivative of tanh|Proposition]]:</strong> $\dfrac{d}{dx}$[[Tanh|$\tanh$]]$(x)=$[[Sech|$\mathrm{sech}$]]$(x)$
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The following formula holds:
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \tanh(z)=\mathrm{sech}^2(z),$$
<strong>Proof:</strong> █
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where $\tanh$ denotes the [[tanh|hyperbolic tangent]] and $\mathrm{sech}$ denotes the [[sech|hyperbolic secant]].
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==Proof==
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From the definition,
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$$\tanh(z) = \dfrac{\sinh(z)}{\cosh(z)},$$
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and so using the [[derivative of sinh]], the [[derivative of cosh]], the [[quotient rule]], the [[Pythagorean identity for sinh and cosh]], and the definition of the [[sech|hyperbolic secant]],
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$$\begin{array}{ll}
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\dfrac{\mathrm{d}}{\mathrm{d}z} \tanh(z) &= \dfrac{\mathrm{d}}{\mathrm{d}z}\left[ \dfrac{\sinh(z)}{\cosh(z)} \right] \\
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&= \dfrac{\cosh^2(z)-\sinh^2(z)}{\cosh^2(z)} \\
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&= \dfrac{1}{\cosh^2(z)} \\
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&= \mathrm{sech}^2(z),
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\end{array}$$
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as was to be shown. █
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==References==
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[[Category:Theorem]]
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[[Category:Proven]]

Latest revision as of 00:02, 17 June 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \tanh(z)=\mathrm{sech}^2(z),$$ where $\tanh$ denotes the hyperbolic tangent and $\mathrm{sech}$ denotes the hyperbolic secant.

Proof

From the definition, $$\tanh(z) = \dfrac{\sinh(z)}{\cosh(z)},$$ and so using the derivative of sinh, the derivative of cosh, the quotient rule, the Pythagorean identity for sinh and cosh, and the definition of the hyperbolic secant, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \tanh(z) &= \dfrac{\mathrm{d}}{\mathrm{d}z}\left[ \dfrac{\sinh(z)}{\cosh(z)} \right] \\ &= \dfrac{\cosh^2(z)-\sinh^2(z)}{\cosh^2(z)} \\ &= \dfrac{1}{\cosh^2(z)} \\ &= \mathrm{sech}^2(z), \end{array}$$ as was to be shown. █

References