Difference between revisions of "Derivative of arctan"

From specialfunctionswiki
Jump to: navigation, search
(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Proposition:</strong> The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arctan}(z) = \dfrac{1...")
 
Line 9: Line 9:
 
[[File:Sec(arctan(z)).png|200px|center]]
 
[[File:Sec(arctan(z)).png|200px|center]]
 
Substituting back in $\theta=\mathrm{arccos(z)}$ yields the formula
 
Substituting back in $\theta=\mathrm{arccos(z)}$ yields the formula
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1}. █$$
+
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1},$$
 +
as was to be shown. █
 
</div>
 
</div>
 
</div>
 
</div>

Revision as of 03:28, 16 May 2016

Proposition: 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 $\sec^2(\mathrm{arctan}(z))=z^2+1$:

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. █