Inverse Trig Function Derivatives

[crayzeerunner]

We are supposed to take the derivative of the function.

The question is: F(t) = [1/(sqrt6)]arctan[((sqrt6)(t))/2]

U = ((sqrt6)t)/2

(1/sqrt6)* 1/(1-(6t/4)) * -1/4
My final answer came to be -1/(4sqrts6) - t
The answer in my book is 1/(2+3t^2)

I believe that I set up my first incorrectly but I do can not find out where/how/why It is wrong. If anyone can help it would be greatly appreciated. Thanks.

 

[skeeter]

derivative of k*arctan(u) = k*u'/(1 + u^2)

F(t) = [1/(sqrt6)]arctan[((sqrt6)(t))/2]

F'(t) = [1/(sqrt6)]*[(sqrt6)/2]/[1 + (6t^2)/4]

F'(t) = (1/2)/[1 + (3t^2)/2]

multiply numerator and denominator by 2 ...

F'(t) = 1/(2 + 3t^2)

 

[crayzeerunner]

Thanks for your help. I was just making a mistake with the derivation of u but I understand now. Thanks again!