Difference between revisions of "Derivative of arctan"
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(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Proposition:</strong> The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arctan}(z) = \dfrac{1...") |
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[[File:Sec(arctan(z)).png|200px|center]] | [[File:Sec(arctan(z)).png|200px|center]] | ||
Substituting back in $\theta=\mathrm{arccos(z)}$ yields the formula | Substituting back in $\theta=\mathrm{arccos(z)}$ yields the formula | ||
| − | $$\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 $\sec^2(\mathrm{arctan}(z))=z^2+1$:
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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. █
