Difference between revisions of "Derivative of sinh"

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==Theorem==
<strong>[[Derivative of sinh|Proposition]]:</strong> The following formula holds:
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The following formula holds:
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z) = \cosh(z),$$
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z) = \cosh(z),$$
 
where $\sinh$ denotes the [[sinh|hyperbolic sine]] and $\cosh$ denotes the [[cosh|hyperbolic cosine]].
 
where $\sinh$ denotes the [[sinh|hyperbolic sine]] and $\cosh$ denotes the [[cosh|hyperbolic cosine]].
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<strong>Proof:</strong> From the definition,
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==Proof==
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From the definition,
 
$$\sinh(z) = \dfrac{e^z-e^{-z}}{2},$$
 
$$\sinh(z) = \dfrac{e^z-e^{-z}}{2},$$
 
and so using the [[derivative of the exponential function]], the [[derivative is a linear operator|linear property of the derivative]], the [[chain rule]], and the definition of the hyperbolic cosine,
 
and so using the [[derivative of the exponential function]], the [[derivative is a linear operator|linear property of the derivative]], the [[chain rule]], and the definition of the hyperbolic cosine,
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z)=\dfrac{e^z + e^{-z}}{2}=\cosh(z),$$
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z)=\dfrac{e^z + e^{-z}}{2}=\cosh(z),$$
 
as was to be shown. █  
 
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 07:52, 8 June 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z) = \cosh(z),$$ where $\sinh$ denotes the hyperbolic sine and $\cosh$ denotes the hyperbolic cosine.

Proof

From the definition, $$\sinh(z) = \dfrac{e^z-e^{-z}}{2},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, and the definition of the hyperbolic cosine, $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z)=\dfrac{e^z + e^{-z}}{2}=\cosh(z),$$ as was to be shown. █

References