I hope I did these right... thanks in advance!

g(x) = (3x-2)/(2x+1)^(1/2)

g'(x) = (9x-1)/(2x+1)^(5/2)

y = sin(cos(tan(x))^(1/2))

y' = [-cos^2(tan(x))^(1/2)sin(tanx)^(1/2)sec^2(x)]/2(tan(x))^(1/2)

these next ones dont need to be simplified

f(x) = sin(x+x^2)

f'(x) = cos(X^(2) + x) * (2x+1)

f(x) = sin((cosx)/x)

f'(x) = cos((cosx)/x) * [ (x(sinx) - cosx)/x^2]

f(x) = sin^3(x^(2)+sinx)

f'(x) = 3[sin(x^(2) + sinx)]^2 * cos(x^(2) + sinx) * (2x + cosx)

f(x) = sin(xsinx) + sin(sin(x^2))

f'(x) = cos(xsinx) * [x(cosx) + (sinx)] + cos(sin(x^(2)) * cos(x^(2)) * 2x

heres the last one...

Find an equation of the tangent line to the curve f(x) = sin(sinx) at the point (pi, 0)

so i found:

f'(x) = cos(sinx)cosx

i plugged that into the y=mx+b as the slope of the tangent line and I got

cos(sin(pi)) * cos(pi) =-1

0=(-1)(pi) + b

b= pi

so, y= -x + pi

Thank you so much for checking this out for me =) I will definitely try to contribute myself around here.