Difference between revisions of "Derivative of tangent"

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==Proof==
 
==Proof==
 
+
From the definition,
 +
$$\tan(z) = \dfrac{\sin(z)}{\cos(z)},$$
 +
so using the [[quotient rule]], the [[derivative of sine]], the [[derivative of cosine]], the [[Pythagorean identity for sin and cos]], and the definition of [[secant]],
 +
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \tan(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{\sin(z)}{\cos(z)} = \dfrac{\cos^2(z) + \sin^2(z)}{\cos^2(z)} = \dfrac{1}{\cos^2(z)} = \sec^2(z),$$
 +
as was to be shown.
 
==References==
 
==References==
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Revision as of 21:23, 21 September 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \tan(z) = \sec^2(z),$$ where $\tan$ denotes the tangent function and $\sec$ denotes the secant function.

Proof

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

References