Inverse Trig Function: Find Derivative of the Function

[chapsticks]

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?

 

[Dick]

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?

What happened to the x??

 

[chapsticks]

is it 4/(|4x|(√4x²-1))

I keep getting it wrong

 

[HallsofIvy]

What is u in your original integral? What is $u^2$?

 

[Dick]

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 $\sqrt(u^2-1)$ 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]

I mean this one (4x)^2

 

[Dick]

I mean this one (4x)^2

Ok, then keep writing it like that. And what about my other question?

 

[chapsticks]

okay how about this answer??

f'(x)=arcsec4x+ 4/(4x(√(16x)²-1)

 

[chapsticks]

I did this one in my homework online and it keeps saying I'm wrong

 

[Dick]

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]

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]

YAY it finally worked thank you :D