Inverse Trig Function: Find Derivative of the Function

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  • #1
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Homework Statement


find the derivative of the function
f(x)=arcsec(4x)


Homework Equations


I think this is a Relevant equations.

d/dx[arcsecu]=u'/(|u|(√u2-1)


The Attempt at a Solution


f'(x)=4/(|4|(√42-1)
=1/√15

I keep getting wrong in my online homework why? :confused:
 

Answers and Replies

  • #2
Dick
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Homework Statement


find the derivative of the function
f(x)=arcsec(4x)


Homework Equations


I think this is a Relevant equations.

d/dx[arcsecu]=u'/(|u|(√u2-1)


The Attempt at a Solution


f'(x)=4/(|4|(√42-1)
=1/√15

I keep getting wrong in my online homework why? :confused:

What happened to the x??
 
  • #3
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is it 4/(|4x|(√4x2-1))

I keep getting it wrong
 
  • #4
HallsofIvy
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What is u in your original integral? What is [itex]u^2[/itex]?
 
  • #5
Dick
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is it 4/(|4x|(√4x2-1))

I keep getting it wrong

That's sort of close. But look up the formula again. Isn't the square root part [itex]\sqrt(u^2-1)[/itex] instead of what you have? And when you write something like 4x^2 it's not clear whether you mean (4x)^2 or 4*(x^2). Which do you mean?
 
  • #6
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I mean this one (4x)^2
 
  • #7
Dick
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I mean this one (4x)^2

Ok, then keep writing it like that. And what about my other question?
 
  • #8
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okay how about this answer??

f'(x)=arcsec4x+ 4/(4x(√(16x)2-1)
 
  • #9
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I did this one in my homework online and it keeps saying I'm wrong
 

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  • #10
Dick
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okay how about this answer??

f'(x)=arcsec4x+ 4/(4x(√(16x)2-1)

Stop changing things without giving any reason. Why did you put the arcsec4x in there? Why did you drop the absolute value on |4x|? (4x)^2 was right, (16x)^2 isn't. Why not?
 
  • #11
Dick
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I did this one in my homework online and it keeps saying I'm wrong

That looks right, except you have x instead of |x|.
 
  • #12
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YAY it finally worked thank you :D
 

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