Difference between revisions of "Derivative of cotangent"
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
| − | $$\dfrac{\mathrm{d}}{\mathrm{d} | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z}\cot(z)=-\csc^2(z),$$ |
where $\cot$ denotes the [[cotangent]] and $\csc$ denotes the [[cosecant]]. | where $\cot$ denotes the [[cotangent]] and $\csc$ denotes the [[cosecant]]. | ||
| Line 7: | Line 7: | ||
Apply the [[quotient rule]] to the definition of [[cotangent]] using [[derivative of sine]] and [[derivative of cosine]] to see | Apply the [[quotient rule]] to the definition of [[cotangent]] using [[derivative of sine]] and [[derivative of cosine]] to see | ||
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
| − | \dfrac{\mathrm{d}}{\mathrm{d} | + | \dfrac{\mathrm{d}}{\mathrm{d}z} \cot(z) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{\cos(z)}{\sin(z)} \right] \\ |
| − | &= \dfrac{-\sin^2( | + | &= \dfrac{-\sin^2(z)-\cos^2(z)}{\sin^2(z)} \\ |
| − | &= -\dfrac{\sin^2( | + | &= -\dfrac{\sin^2(z)+\cos^2(z)}{\sin^2(z)}. |
\end{array}$$ | \end{array}$$ | ||
Now apply the [[Pythagorean identity for sin and cos]] and the definition of [[cosecant]] to see | Now apply the [[Pythagorean identity for sin and cos]] and the definition of [[cosecant]] to see | ||
| − | $$\dfrac{\mathrm{d}}{\mathrm{d} | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \cot(z) = -\dfrac{1}{\sin^2(z)} = -\csc^2(z),$$ |
| − | as was to be shown. █ | + | as was to be shown. █ |
==References== | ==References== | ||
Latest revision as of 01:29, 1 July 2017
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z}\cot(z)=-\csc^2(z),$$ where $\cot$ denotes the cotangent and $\csc$ denotes the cosecant.
Proof
Apply the quotient rule to the definition of cotangent using derivative of sine and derivative of cosine to see $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \cot(z) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{\cos(z)}{\sin(z)} \right] \\ &= \dfrac{-\sin^2(z)-\cos^2(z)}{\sin^2(z)} \\ &= -\dfrac{\sin^2(z)+\cos^2(z)}{\sin^2(z)}. \end{array}$$ Now apply the Pythagorean identity for sin and cos and the definition of cosecant to see $$\dfrac{\mathrm{d}}{\mathrm{d}z} \cot(z) = -\dfrac{1}{\sin^2(z)} = -\csc^2(z),$$ as was to be shown. █
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.3.110$
