Difference between revisions of "Derivative of arccos"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}},$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}},$$ | ||
where $\mathrm{arccos}$ denotes the [[arccos|inverse cosine]] function. | where $\mathrm{arccos}$ denotes the [[arccos|inverse cosine]] function. | ||
| − | + | ||
| − | + | ==Proof== | |
| + | If $\theta=\mathrm{arccos}(z)$ then $\cos(\theta)=z$. Now use [[implicit differentiation]] with respect to $z$ to get | ||
$$-\sin(\theta)\theta'=1.$$ | $$-\sin(\theta)\theta'=1.$$ | ||
The following image shows that $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$: <br /> | The following image shows that $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$: <br /> | ||
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Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula <br /> | Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula <br /> | ||
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$ | ||
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 07:29, 8 June 2016
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}},$$ where $\mathrm{arccos}$ denotes the inverse cosine function.
Proof
If $\theta=\mathrm{arccos}(z)$ then $\cos(\theta)=z$. Now use implicit differentiation with respect to $z$ to get
$$-\sin(\theta)\theta'=1.$$
The following image shows that $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$:
).png)
Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$
