Difference between revisions of "Derivative of cotangent"

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==Theorem==
<strong>[[Derivative of cotangent|Proposition]]:</strong> The following formula holds:
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The following formula holds:
 
$$\dfrac{\mathrm{d}}{\mathrm{d}x}\cot(x)=-\csc^2(x),$$
 
$$\dfrac{\mathrm{d}}{\mathrm{d}x}\cot(x)=-\csc^2(x),$$
 
where $\cot$ denotes the [[cotangent]] and $\csc$ denotes the [[cosecant]].
 
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
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==Proof==
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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}{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] \\
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$$\dfrac{d}{dx} \cot(x) = -\dfrac{1}{\sin^2(x)} = -\csc^2(x),$$
 
$$\dfrac{d}{dx} \cot(x) = -\dfrac{1}{\sin^2(x)} = -\csc^2(x),$$
 
as was to be shown. █  
 
as was to be shown. █  
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==References==
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[[Category:Theorem]]
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[[Category:Proven]]

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 $$\begin{array}{ll} \dfrac{d}{dx} \cot(x) &= \dfrac{d}{dx} \left[ \dfrac{\cos(x)}{\sin(x)} \right] \\ &= \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} \\ &= -\dfrac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}. \end{array}$$ 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. █

References