Difference between revisions of "Derivative of arcsec"

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==Theorem==
<strong>[[Derivative of arcsec|Theorem]]:</strong> The following formula holds:
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The following formula holds:
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsec}(z) = \dfrac{1}{z^2\sqrt{1-\frac{1}{z^2}}},$$
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsec}(z) = \dfrac{1}{z^2\sqrt{1-\frac{1}{z^2}}},$$
 
where $\mathrm{arcsec}$ is the [[arcsec|inverse secant]] function.
 
where $\mathrm{arcsec}$ is the [[arcsec|inverse secant]] function.
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<strong>Proof:</strong> █
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==Proof==
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If $\theta=\mathrm{arcsec}(z)$ then $\sec(\theta)=z$. Now use [[implicit differentiation]] with respect to $z$ and the [[derivative of secant]] to see
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$$\sec(\theta)\tan(\theta) \theta' = 1,$$
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or equivalently,
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$$\dfrac{\mathrm{d}\theta}{\mathrm{d}z} = \dfrac{1}{\sec(\theta)\tan(\theta)} = \dfrac{1}{z\tan(\theta)}.$$
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The following image shows that $\tan(\mathrm{arcsec}(z))=\sqrt{z^2-1}$:
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[[File:Tan(arcsec(z)).png|200px|center]]
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Hence substituting back in $\theta=\mathrm{arcsec}(z)$ yields the formula
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsec}(z) = \dfrac{1}{z\tan(\mathrm{arcsec}(z))} = \dfrac{1}{z\sqrt{z^2-1}}=\dfrac{1}{z^2\sqrt{1-\frac{1}{z^2}}},$$
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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 23:51, 8 December 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsec}(z) = \dfrac{1}{z^2\sqrt{1-\frac{1}{z^2}}},$$ where $\mathrm{arcsec}$ is the inverse secant function.

Proof

If $\theta=\mathrm{arcsec}(z)$ then $\sec(\theta)=z$. Now use implicit differentiation with respect to $z$ and the derivative of secant to see $$\sec(\theta)\tan(\theta) \theta' = 1,$$ or equivalently, $$\dfrac{\mathrm{d}\theta}{\mathrm{d}z} = \dfrac{1}{\sec(\theta)\tan(\theta)} = \dfrac{1}{z\tan(\theta)}.$$ The following image shows that $\tan(\mathrm{arcsec}(z))=\sqrt{z^2-1}$:

Tan(arcsec(z)).png

Hence substituting back in $\theta=\mathrm{arcsec}(z)$ yields the formula $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsec}(z) = \dfrac{1}{z\tan(\mathrm{arcsec}(z))} = \dfrac{1}{z\sqrt{z^2-1}}=\dfrac{1}{z^2\sqrt{1-\frac{1}{z^2}}},$$ as was to be shown.

References