Difference between revisions of "Derivative of cotangent"
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$$\begin{array}{ll} | $$\begin{array}{ll} | ||
\dfrac{d}{dx} \cot(x) &= \dfrac{d}{dx} \left[ \dfrac{\cos(x)}{\sin(x)} \right] \\ | \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)} \\ |
| + | &= -\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 | ||
Revision as of 18:29, 1 December 2015
Proposition: $\dfrac{d}{dx}$$\cot$$(x)=-$$\csc$$^2(x)$
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. █
