Quote:
Originally Posted by YouAreWet
Okay, basic calculus here. I'm stumped trying to study for a test tomorrow and, when in doubt, 2+2 thread.
f(x) = 2arcsin(x-1)
I know the formula for arcsinu is u(prime) / (squareroot of: 1- u^2)
So, u = x-1 in this case.
Differentiate: 2 * 1 / (squareroot of: 1 - (x-1)^2)
How do I continue from here?
I have an answer from my book of 2 / (squareroot of: 2x - x^2) but I can't seem to get there. Maybe my question is more of basic algebra the more that I think about it.
This is just algebra:
2*1/ (sqrt(1-(x-1)^2) = 2/sqrt(1 - (x^2 -2x +1)) = 2/(2x - x^2)
by the way, the derivative wrt x of arcsin(x) = 1/(sqrt(1-x^2)
If you don't want to memorize, then you can do it the longer way.
replace f(x) with y.
y=2*arcsin(x-1)
y/2 = arcsin(x-1)
sin(y/2) = (x-1) by definition
Take the derivative of both sides wrt x
cos(y/2)*dy/dx = 1
dy/dx = 1/cos(y/2)
we know sin^2(y/2) + cos^2(y/2) = 1
or equivalently,
cos^2(y/2) = 1 - sin^2(y/2)
cos^2(y/2) = sqrt(1-sin^2(y/2))
since we know sin(y/2) = x-1 , then we have
cos^2(y/2) = 1 - sin^2(y/2) = 1 - (x-1)^2 = 2x - x^2
therefore,
dy/dx = 1/(2x - x^2)
