Difference between revisions of "Arctan"
From specialfunctionswiki
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$$\dfrac{d}{dz} \mathrm{arctan}(z) = \dfrac{1}{z^2+1}$$ | $$\dfrac{d}{dz} \mathrm{arctan}(z) = \dfrac{1}{z^2+1}$$ | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
| − | <strong>Proof:</strong> █ | + | <strong>Proof:</strong> If $y=\mathrm{arctan}(z)$ then $\tan y = z$. Now use [[implicit differentiation]] with respect to $z$ yields |
| + | $$\sec^2(y)y'=1.$$ | ||
| + | Substituting back in $y=\mathrm{arccos(z)}$ yields the formula | ||
| + | $$\dfrac{d}{dz} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1}. █$$ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 02:00, 28 October 2014
The $\mathrm{arctan}$ function is the inverse function of the tangent function.
Properties
Proposition: $$\dfrac{d}{dz} \mathrm{arctan}(z) = \dfrac{1}{z^2+1}$$
Proof: If $y=\mathrm{arctan}(z)$ then $\tan y = z$. Now use implicit differentiation with respect to $z$ yields $$\sec^2(y)y'=1.$$ Substituting back in $y=\mathrm{arccos(z)}$ yields the formula $$\dfrac{d}{dz} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1}. █$$
Proposition: $$\int \mathrm{arctan}(z) = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C$$
Proof: █
Proposition: $$\mathrm{arctan}(z) = \mathrm{arccot}\left( \dfrac{1}{z} \right)$$
Proof: █
