Difference between revisions of "Derivative of arcsin"
From specialfunctionswiki
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\sqrt{1-z^2}},$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\sqrt{1-z^2}},$$ | ||
where $\arcsin$ denotes the [[arcsin|inverse sine]] function. | where $\arcsin$ denotes the [[arcsin|inverse sine]] function. | ||
| − | + | ||
| − | + | ==Proof== | |
| + | If $\theta=\mathrm{arcsin}(z)$ then $\sin(\theta)=z$. Now use [[implicit differentiation]] with respect to $z$ to get | ||
$$\cos(\theta)\theta'=1.$$ | $$\cos(\theta)\theta'=1.$$ | ||
The following image shows that $\cos(\mathrm{arcsin}(z))=\sqrt{1-z^2}$: | The following image shows that $\cos(\mathrm{arcsin}(z))=\sqrt{1-z^2}$: | ||
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\cos(\mathrm{arcsin(z)})} = \dfrac{1}{\sqrt{1-z^2}},$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\cos(\mathrm{arcsin(z)})} = \dfrac{1}{\sqrt{1-z^2}},$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
Revision as of 07:20, 8 June 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. █
