Difference between revisions of "Derivative of secant"

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==Proof==
 
==Proof==
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From the definition of secant,
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$$\sec(z) = \dfrac{1}{\cos(z)},$$
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and so using the [[quotient rule]], the [[derivative of cosine]], and the definition of [[tangent]],
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sec(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{1}{\cos(z)} = \dfrac{\sin(z)}{\cos^2(z)}=\tan(z)\sec(z),$$
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as was to be shown. $\blacksquare$
  
 
==References==
 
==References==
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*{{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Derivative of cosecant|next=Derivative of cotangent}}: $4.3.109$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
[[Category:Unproven]]
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[[Category:Proven]]

Latest revision as of 00:34, 26 April 2017

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sec(z)=\tan(z)\sec(z),$$ where $\sec$ denotes the secant and $\cot$ denotes the cotangent.

Proof

From the definition of secant, $$\sec(z) = \dfrac{1}{\cos(z)},$$ and so using the quotient rule, the derivative of cosine, and the definition of tangent, $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sec(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{1}{\cos(z)} = \dfrac{\sin(z)}{\cos^2(z)}=\tan(z)\sec(z),$$ as was to be shown. $\blacksquare$

References