Difference between revisions of "Derivative of cosecant"

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\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] \\
 
&= \dfrac{0-\cos(z)}{\sin^2(z)} \\
 
&= \dfrac{0-\cos(z)}{\sin^2(z)} \\
&= \csc(z)\cot(z),
+
&= -\csc(z)\cot(z),
 
\end{array}$$
 
\end{array}$$
 
as was to be shown. █  
 
as was to be shown. █  
 
</div>
 
</div>
 
</div>
 
</div>

Revision as of 04:28, 8 February 2016

Proposition: 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. █