Difference between revisions of "Derivative of arctan"

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==Theorem==
<strong>Proposition:</strong> The following formula holds:
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The following formula holds:
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arctan}(z) = \dfrac{1}{z^2+1},$$
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arctan}(z) = \dfrac{1}{z^2+1},$$
 
where $\mathrm{arctan}$ denotes the [[arctan|inverse tangent]] function.
 
where $\mathrm{arctan}$ denotes the [[arctan|inverse tangent]] function.
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<strong>Proof:</strong> If $\theta=\mathrm{arctan}(z)$ then $\tan \theta = z$. Now use [[implicit differentiation]] with respect to $z$ yields
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==Proof==
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If $\theta=\mathrm{arctan}(z)$ then $\tan \theta = z$. Now [[implicit differentiation]] with respect to $z$ yields
 
$$\sec^2(\theta)\theta'=1.$$
 
$$\sec^2(\theta)\theta'=1.$$
 
The following triangle shows that $\sec^2(\mathrm{arctan}(z))=z^2+1$:
 
The following triangle shows that $\sec^2(\mathrm{arctan}(z))=z^2+1$:
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1},$$  
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1},$$  
 
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 13:05, 22 September 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arctan}(z) = \dfrac{1}{z^2+1},$$ where $\mathrm{arctan}$ denotes the inverse tangent function.

Proof

If $\theta=\mathrm{arctan}(z)$ then $\tan \theta = z$. Now implicit differentiation with respect to $z$ yields $$\sec^2(\theta)\theta'=1.$$ The following triangle shows that $\sec^2(\mathrm{arctan}(z))=z^2+1$:

Sec(arctan(z)).png

Substituting back in $\theta=\mathrm{arccos(z)}$ yields the formula $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1},$$ as was to be shown. █

References