Recent content by mleeno

AH so simply f'(pi/6) = g''(pi/6) thanks for the help everyone. I got something like f'(x) = (cosx)sqrt(1 + (sinx)^2) plugging in f'(pi/6) = cos(pi/6)sqrt(1 + (sin(pi/6)^2)) which should be [sqrt(3)/2]*sqrt(5/4) ?

g'(y) = f(y) ? g'(y) = f(y)y' I think I'm thinking about this too hard...

Homework Statement problem in attachment Homework Equations The Attempt at a Solution I can get f'(x) as sqrt(1 + (sinx)^2) and derive that to get the second derivative but as far as that I don't really get the concept behind this question will y be another function I have...