Difference between revisions of "Arccot"

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(Properties)
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=Properties=
 
=Properties=
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{{:Derivative of arccot}}
<strong>Proposition:</strong>
 
$$\dfrac{d}{dz} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1}$$
 
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<strong>Proof:</strong> If $y=\mathrm{arccot}(z)$ then $\cot(y)=z$. Now use [[implicit differentiation]] with respect to $z$ to get
 
$$-\csc^2(y)y'=1.$$
 
Substituting back in $y=\mathrm{arccos}(z)$ yields the formula
 
$$\dfrac{d}{dz} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1}.█$$ 
 
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=References=
 
=References=

Revision as of 21:35, 15 May 2016

There are two functions commonly called $\mathrm{arccot}$, which refers to inverse functions of the $\mathrm{cot}$ function. First is the function $\mathrm{arccot_1}\colon \mathbb{R} \rightarrow (0,\pi)$ which results from restricting cotangent to $(0,\pi)$ and second is the function $\mathrm{arccot_2} \colon \mathbb{R} \rightarrow \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \setminus \{0\}$ which results from restricting cotangent to $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.

Properties

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1},$$ where $\mathrm{arccot}$ denotes the inverse cotangent function.

Proof

If $y=\mathrm{arccot}(z)$ then $\cot(y)=z$. Now use implicit differentiation with respect to $z$ to get $$-\csc^2(y)y'=1.$$ Substituting back in $y=\mathrm{arccos}(z)$ yields the formula $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1},$$ as was to be shown. █

References

References

Which is the correct graph of arccot x?

See Also

Cotangent
Coth
Arccoth

<center>Inverse trigonometric functions
</center>