Difference between revisions of "Derivative of cotangent"

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==Theorem==
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<strong>[[Derivative of cotangent|Proposition]]:</strong> $\dfrac{d}{dx}$[[Cotangent|$\cot$]]$(x)=-$[[Cosecant|$\csc$]]$^2(x)$
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The following formula holds:
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$$\dfrac{\mathrm{d}}{\mathrm{d}z}\cot(z)=-\csc^2(z),$$
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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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where $\cot$ denotes the [[cotangent]] and $\csc$ denotes the [[cosecant]].
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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}
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\dfrac{d}{dx} \cot(x) &= \dfrac{d}{dx} \left[ \dfrac{\cos(x)}{\sin(x)} \right] \\
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\dfrac{\mathrm{d}}{\mathrm{d}z} \cot(z) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{\cos(z)}{\sin(z)} \right] \\
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&= \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)}.
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&= \dfrac{-\sin^2(z)-\cos^2(z)}{\sin^2(z)} \\
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&= -\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
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$$\dfrac{d}{dx} \cot(x) = -\dfrac{1}{\sin^2(x)} = -\csc^2(x),$$
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \cot(z) = -\dfrac{1}{\sin^2(z)} = -\csc^2(z),$$
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as was to be shown. █  
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as was to be shown. █
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</div>
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</div>
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==References==
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*{{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Derivative of secant|next=findme}}: $4.3.110$
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[[Category:Theorem]]
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[[Category:Proven]]

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

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