Difference between revisions of "Derivative of arcsin"
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(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{...") |
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[[File:Cos(arcsin(z)).png|200px|center]] | [[File:Cos(arcsin(z)).png|200px|center]] | ||
Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula | Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula | ||
| − | $$\dfrac{d}{ | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\cos(\mathrm{arcsin(z)})} = \dfrac{1}{\sqrt{1-z^2}},$$ |
as was to be shown. █ | as was to be shown. █ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 21:16, 15 May 2016
Theorem: The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\sqrt{1-z^2}},$$ where $\arcsin$ denotes the inverse sine function.
Proof: If $\theta=\mathrm{arcsin}(z)$ then $\sin(\theta)=z$. Now use implicit differentiation with respect to $z$ to get $$\cos(\theta)\theta'=1.$$ The following image shows that $\cos(\mathrm{arcsin}(z))=\sqrt{1-z^2}$:
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Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\cos(\mathrm{arcsin(z)})} = \dfrac{1}{\sqrt{1-z^2}},$$ as was to be shown. █
