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What is the derivative of #y= ln abs(secx-tanx)# for #x>0#?

Answer 1

Isabella Ackerman

#d/dx(ln abs(secx-tanx)) = -secx #

Using the chain rule:

#d/dx(ln abs(secx-tanx)) = 1/(secx-tanx) d/dx (secx-tanx)#

#d/dx(ln abs(secx-tanx)) = (secxtanx-sec^2x)/(secx-tanx) #

#d/dx(ln abs(secx-tanx)) = (secx(tanx-secx))/(secx-tanx) #

#d/dx(ln abs(secx-tanx)) = -secx #

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Answer 2

Axel Alvarez

The derivative of ( y = \ln|\sec(x) - \tan(x)| ) for ( x > 0 ) is:

[ y' = \frac{d}{dx}(\ln|\sec(x) - \tan(x)|) = \frac{\sec(x)\tan(x) - \sec(x)}{\sec(x) - \tan(x)} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What is the derivative of #y= ln abs(secx-tanx)# for #x&gt;0#?

What is the derivative of #y= ln abs(secx-tanx)# for #x>0#?