Hw do you prove that the d/dx (secx) = sec x tan x ?

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Answer 1 · 2017-10-22T22:06:40

Use the Quotient rule

STEP 1:
$sec x = \frac{1}{cos x}$ $sec x = 1 / cos x$

STEP 2:
Use the quotient rule to find the derivative of $sec x = \frac{1}{cos x}$ $Sec x = 1 / cos x$

$\frac{\frac{d}{d x} 1 \cdot (cos x) - \frac{d}{d x} cos x \cdot (1)}{{(cos x)}^{2}}$ $(d/dx 1 * (cos x) - d/dx cosx * (1)) / (cosx)^2$

STEP 3:
Simplify

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

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

$= (\frac{sin x}{cos x}) (\frac{1}{cos x})$ $= (sinx / cos x ) (1 / cosx)$

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