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Inverse Trig Function: Find Derivative of the Function

by chapsticks
Tags: derivative, function, inverse, trig
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chapsticks
#1
Jan19-12, 10:38 PM
P: 38
1. The problem statement, all variables and given/known data
find the derivative of the function
f(x)=arcsec(4x)


2. Relevant equations
I think this is a Relevant equations.

d/dx[arcsecu]=u'/(|u|(√u2-1)


3. The attempt at a solution
f'(x)=4/(|4|(√42-1)
=1/√15

I keep getting wrong in my online homework why?
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Dick
#2
Jan19-12, 10:48 PM
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Quote Quote by chapsticks View Post
1. The problem statement, all variables and given/known data
find the derivative of the function
f(x)=arcsec(4x)


2. Relevant equations
I think this is a Relevant equations.

d/dx[arcsecu]=u'/(|u|(√u2-1)


3. The attempt at a solution
f'(x)=4/(|4|(√42-1)
=1/√15

I keep getting wrong in my online homework why?
What happened to the x??
chapsticks
#3
Jan20-12, 03:07 PM
P: 38
is it 4/(|4x|(√4x2-1))

I keep getting it wrong

HallsofIvy
#4
Jan20-12, 03:20 PM
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PF Gold
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Inverse Trig Function: Find Derivative of the Function

What is u in your original integral? What is [itex]u^2[/itex]?
Dick
#5
Jan20-12, 03:25 PM
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Quote Quote by chapsticks View Post
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?
chapsticks
#6
Jan20-12, 03:27 PM
P: 38
I mean this one (4x)^2
Dick
#7
Jan20-12, 03:32 PM
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Quote Quote by chapsticks View Post
I mean this one (4x)^2
Ok, then keep writing it like that. And what about my other question?
chapsticks
#8
Jan20-12, 03:43 PM
P: 38
okay how about this answer??

f'(x)=arcsec4x+ 4/(4x(√(16x)2-1)
chapsticks
#9
Jan20-12, 03:49 PM
P: 38
I did this one in my homework online and it keeps saying I'm wrong
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Dick
#10
Jan20-12, 03:52 PM
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Quote Quote by chapsticks View Post
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?
Dick
#11
Jan20-12, 03:54 PM
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Quote Quote by chapsticks View Post
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|.
chapsticks
#12
Jan20-12, 03:57 PM
P: 38
YAY it finally worked thank you :D


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