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What is the derivative of # y = ( sec x + tan x )( sec x - tan x )#?

Answer 1

Elijah Abram

#y'=0#

#y=(secx+tanx)(secx-tanx)=sec^2x-tan^2x=1#

#y'=0#

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Answer 2

Ezra Adams

To find the derivative of ( y = (\sec x + \tan x)(\sec x - \tan x) ), you can use the product rule.

The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), the derivative of their product ( u(x) \cdot v(x) ) is given by ( u'(x)v(x) + u(x)v'(x) ).

Let ( u(x) = \sec x + \tan x ) and ( v(x) = \sec x - \tan x ).

The derivatives are ( u'(x) = \sec x \tan x + \sec^2 x ) and ( v'(x) = \sec x \tan x - \sec^2 x ).

Now, apply the product rule:

[ y' = u'(x)v(x) + u(x)v'(x) ] [ = (\sec x \tan x + \sec^2 x)(\sec x - \tan x) + (\sec x + \tan x)(\sec x \tan x - \sec^2 x) ]

This simplifies to:

[ y' = \sec^3 x - \tan^3 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.