Derivatives of Inverse Trig Functions

[Guest]

f(x)=arcsec(2x) find the derivative

my work........

rule.. the derivative of the arcsec (u(x)) is 1/abs(u(x))*sqrt (u^2-1)

so, dy/dx= 2/abs(2x)*sqrt(4x^2-1)

However, when I put this into my TI89, it presents a different answer. Is this correct, and if not what mistake is there.

Also, y=25arcsin(x/5)-x sqrt(25-x^2)

dy/dx= 5/sqrt(1-(x/5)^2)-[-x^2/sqrt(25-x^2)+ sqrt(25-x^2)

again my calculator gave me a different answer.

Any help is appreciated.

 

[arthur ohlsten]

y=sec^-1 [2x] take sec of both sides
sec y =2x but secy = 1/cos y this isn't neccessary, but I find it easier
1/ cos y =2x take derivative of each side
[siny/ cos^2 y] [dy/dx] =2
dy/dx = 2 cos^2 y / siny

but y= sec^-1 2x or
sec y =2x or
1/cos y =2x or
cos y = 1/[2x]
sketch a right triangle with the angle of y, adjacent side 1, and hypoteneuse 2x
and opposite side sq rt [4x^2-1] this satisfys the above equation

then cos^2 y =1/[4x^2]
sin y = [4x^2-1]^1/2 /[2x]
substituting
dy/dx = 2 cos^2 y /sin y
2[2x]
dy/dx = -----------------------
[4x^2][4x^2-1]^1/2

dy/dx = 1 /{ x [4x^2-1]^1/2 }
same answer as you got