Difference between revisions of "Derivative of secant"
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and so using the [[quotient rule]], the [[derivative of cosine]], and the definition of [[tangent]], | and so using the [[quotient rule]], the [[derivative of cosine]], and the definition of [[tangent]], | ||
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sec(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{1}{\cos(z)} = \dfrac{\sin(z)}{\cos^2(z)}=\tan(z)\sec(z),$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sec(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{1}{\cos(z)} = \dfrac{\sin(z)}{\cos^2(z)}=\tan(z)\sec(z),$$ | ||
| − | as was to be shown. | + | as was to be shown. $\blacksquare$ |
==References== | ==References== | ||
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
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.3.109$
