Difference between revisions of "Derivative of cotangent"
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| − | <strong>[[Derivative of cotangent|Proposition]]:</strong> $\dfrac{d}{dx}$[[Cotangent|$\cot$]]$(x)=-$[[Cosecant|$\csc$]]$^2(x)$ | + | <strong>[[Derivative of cotangent|Proposition]]:</strong> The following formula holds: |
| + | $$\dfrac{d}{dx}$[[Cotangent|$\cot$]]$(x)=-$[[Cosecant|$\csc$]]$^2(x),$$ | ||
| + | where $\cot$ denotes the [[cotangent]] and $\csc$ denotes the [[cosecant]]. | ||
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<strong>Proof:</strong> Apply the [[quotient rule]] to the definition of [[cotangent]] using [[derivative of sine]] and [[derivative of cosine]] to see | <strong>Proof:</strong> Apply the [[quotient rule]] to the definition of [[cotangent]] using [[derivative of sine]] and [[derivative of cosine]] to see | ||
Revision as of 05:03, 8 February 2016
Proposition: The following formula holds: $$\dfrac{d}{dx}$$\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 $$\begin{array}{ll} \dfrac{d}{dx} \cot(x) &= \dfrac{d}{dx} \left[ \dfrac{\cos(x)}{\sin(x)} \right] \\ &= \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} \\ &= -\dfrac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}. \end{array}$$ Now apply the Pythagorean identity for sin and cos and the definition of cosecant to see $$\dfrac{d}{dx} \cot(x) = -\dfrac{1}{\sin^2(x)} = -\csc^2(x),$$ as was to be shown. █
