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. █