Difference between revisions of "Derivative of cosecant"

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==Theorem==
<strong>[[Derivative of cosecant|Proposition]]:</strong> The following formula holds:
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The following formula holds:
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \csc(z)=- \cot(z)\csc(z),$$
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \csc(z)=- \cot(z)\csc(z),$$
 
where $\csc$ denotes the [[cosecant]] function and $\cot$ denotes the [[cotangent]] function.
 
where $\csc$ denotes the [[cosecant]] function and $\cot$ denotes the [[cotangent]] function.
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<strong>Proof:</strong> Using the [[product rule]] and the definitions of [[cosecant]] and [[cotangent]],  
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==Proof==
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Using the [[product rule]] and the definitions of [[cosecant]] and [[cotangent]],  
 
$$\begin{array}{ll}
 
$$\begin{array}{ll}
 
\dfrac{\mathrm{d}}{\mathrm{d}z} \csc(z) &= \dfrac{\mathrm{d}}{\mathrm{d}z} \left[ \dfrac{1}{\sin(z)} \right] \\
 
\dfrac{\mathrm{d}}{\mathrm{d}z} \csc(z) &= \dfrac{\mathrm{d}}{\mathrm{d}z} \left[ \dfrac{1}{\sin(z)} \right] \\
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\end{array}$$
 
\end{array}$$
 
as was to be shown. █  
 
as was to be shown. █  
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==References==

Revision as of 04:35, 6 June 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \csc(z)=- \cot(z)\csc(z),$$ where $\csc$ denotes the cosecant function and $\cot$ denotes the cotangent function.

Proof

Using the product rule and the definitions of cosecant and cotangent, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \csc(z) &= \dfrac{\mathrm{d}}{\mathrm{d}z} \left[ \dfrac{1}{\sin(z)} \right] \\ &= \dfrac{0-\cos(z)}{\sin^2(z)} \\ &= -\csc(z)\cot(z), \end{array}$$ as was to be shown. █

References