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Question #c181e

1 Answer

Oct 25, 2016

Here is a proof using the quotient rule and $\frac{d}{d x} (cos x) = - sin x$ $d/dx(cosx) = -sinx$

Explanation:

$\frac{d}{d x} (sec x) = \frac{d}{d x} (\frac{1}{cos x})$ $d/dx(secx) = d/dx(1/cosx)$

$= \frac{(0) (cos x) - (1) (- sin x)}{{(cos x)}^{2}}$ $= ((0)(cosx)-(1)(-sinx))/(cosx)^2$

$= \frac{sin x}{{(cos x)}^{2}}$ $= sinx/(cosx)^2$

$= \frac{1}{cos x} \cdot \frac{sin x}{cos x}$ $= 1/cosx * sinx/cosx$

$= sec x tan x$ $= secx tanx$

Here's a link to a proof using the definition of derivative

https://socratic.org/questions/how-do-you-use-limit-to-prove-the-derivative-of-secx-secxtanx#132218

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See all questions in Derivative Rules for y=cos(x) and y=tan(x)

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