Difference between revisions of "Derivative of cotangent"

From specialfunctionswiki
Jump to: navigation, search
(Proof)
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
==Theorem==
 
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$\dfrac{\mathrm{d}}{\mathrm{d}x}\cot(x)=-\csc^2(x),$$
+
$$\dfrac{\mathrm{d}}{\mathrm{d}z}\cot(z)=-\csc^2(z),$$
 
where $\cot$ denotes the [[cotangent]] and $\csc$ denotes the [[cosecant]].
 
where $\cot$ denotes the [[cotangent]] and $\csc$ denotes the [[cosecant]].
  
Line 7: Line 7:
 
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{\mathrm{d}}{\mathrm{d}x} \cot(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{\cos(x)}{\sin(x)} \right] \\
+
\dfrac{\mathrm{d}}{\mathrm{d}z} \cot(z) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{\cos(z)}{\sin(z)} \right] \\
&= \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} \\
+
&= \dfrac{-\sin^2(z)-\cos^2(z)}{\sin^2(z)} \\
&= -\dfrac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}.
+
&= -\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
$$\dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) = -\dfrac{1}{\sin^2(x)} = -\csc^2(x),$$
+
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \cot(z) = -\dfrac{1}{\sin^2(z)} = -\csc^2(z),$$
as was to be shown. █  
+
as was to be shown. █
  
 
==References==
 
==References==

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