Difference between revisions of "Derivative of arctan"
From specialfunctionswiki
| Line 1: | Line 1: | ||
| − | + | ==Theorem== | |
| − | + | 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. | ||
| − | + | ||
| − | + | ==Proof== | |
| + | If $\theta=\mathrm{arctan}(z)$ then $\tan \theta = z$. Now use [[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 $\sec^2(\mathrm{arctan}(z))=z^2+1$: | ||
| Line 11: | Line 12: | ||
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1},$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1},$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
Revision as of 07:25, 8 June 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 use implicit differentiation with respect to $z$ yields $$\sec^2(\theta)\theta'=1.$$ The following triangle shows that $\sec^2(\mathrm{arctan}(z))=z^2+1$:
).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. █
