Proving Horizontal Tangents of y=Cos(x) & y=Sec(x) at x=0

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Homework Help Overview

The discussion revolves around proving that the functions y = Cos(x) and y = Sec(x) have horizontal tangents at x = 0. Participants are exploring the derivatives of these functions and the conditions under which horizontal tangents occur.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss their attempts to differentiate the functions and evaluate the derivatives at x = 0. There is a focus on understanding the implications of the product sec(x)tan(x) and the conditions for it to equal zero.

Discussion Status

Some participants express confusion about the reasoning behind setting factors equal to zero in the context of the derivative. Others provide clarification on the evaluation of the derivative at x = 0, noting that both sec(0) and tan(0) contribute to the result being zero. The discussion reflects a mix of interpretations and attempts to clarify the mathematical reasoning involved.

Contextual Notes

Participants are navigating the challenge of proving horizontal tangents while grappling with the definitions and behaviors of trigonometric functions at specific points. There is an emphasis on understanding the conditions under which derivatives yield horizontal tangents.

Robokapp
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im supposed to prove that

y=Cos(x)

and

y=Sec(x)

have horisontal tangents for x=0

i got the derivatives just fine, i proved the first on e with no problem, but what do i do with second? i get to a point where i must prove that Sec(x)Tan(x)=0 but sec x = 1/cos(x) and i can't work with that...i mean how can i divide 1 by something to get zero? it's an asymptote...so what do i do?

i know how to get derivatives, i don't know how to prove that the second one is having the horisontal tangent.
 
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y = secx
y' = secx.tanx = sinx/cos²x

at x = 0, sinx = 0, cosx = 1

y' = 0/1² = 0
==========
 
Fermat said:
y = secx
y' = secx.tanx = sinx/cos²x

at x = 0, sinx = 0, cosx = 1

y' = 0/1² = 0
==========

okay...i thought that when you have secx*tanx=0 you have to set them like when you have parenthesis...each one at a time. That was probably a better explanation than most teachers would give. Thank you.

What i was trying to do is prove that either tan or sec is some value and the second one is zero, so zero*value=0 but it doesn't work that way.

can i ask though...why was i wrong? i mean if you do (x)(X+1)=0 you're defenetly not wrong to set either one equl to zero...why wasn't my way working? (To be honest I am more interested on what went wrong than what the answer is...i can always copy that from the book but i don't want to).
 
As far as I can see, you weren't wrong!

You had,

y' = secx*tanx

at x = 0, secx= 1/cosx = 1/1 = 1, and tanx = 0,

So,

y' = 1*0 = 0

what you have is secx = 1 and the other one, tanx = 0
 

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