Difference between revisions of "Derivative of arccot"
From specialfunctionswiki
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1},$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1},$$ | ||
where $\mathrm{arccot}$ denotes the [[arccot|inverse cotangent]] function. | where $\mathrm{arccot}$ denotes the [[arccot|inverse cotangent]] function. | ||
| − | + | ||
| − | + | ==Proof== | |
| + | If $y=\mathrm{arccot}(z)$ then $\cot(y)=z$. Now use [[implicit differentiation]] with respect to $z$ to get | ||
$$-\csc^2(y)y'=1.$$ | $$-\csc^2(y)y'=1.$$ | ||
Substituting back in $y=\mathrm{arccos}(z)$ yields the formula | Substituting back in $y=\mathrm{arccos}(z)$ yields the formula | ||
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1},$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1},$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 07:31, 8 June 2016
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. █
