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.