Difference between revisions of "Derivative of cotangent"
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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{d}{ | + | \dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{\cos(x)}{\sin(x)} \right] \\ |
&= \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} \\ | &= \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} \\ | ||
&= -\dfrac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}. | &= -\dfrac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}. | ||
\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 | ||
| − | $$\dfrac{d}{ | + | $$\dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) = -\dfrac{1}{\sin^2(x)} = -\csc^2(x),$$ |
as was to be shown. █ | as was to be shown. █ | ||
Revision as of 07:47, 8 June 2016
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x}\cot(x)=-\csc^2(x),$$ where $\cot$ denotes the cotangent and $\csc$ denotes the cosecant.
Proof
Apply the quotient rule to the definition of cotangent using derivative of sine and derivative of cosine to see $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{\cos(x)}{\sin(x)} \right] \\ &= \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} \\ &= -\dfrac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}. \end{array}$$ Now apply the Pythagorean identity for sin and cos and the definition of cosecant to see $$\dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) = -\dfrac{1}{\sin^2(x)} = -\csc^2(x),$$ as was to be shown. █
