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.
It was just the tan(x) I think I typed it it wrong =( sorry
Edit: I think I typed it right? It was supposed to be just the tan(x) in the square root.
I had the minus sign, I mistyped again =(
I don't know what that means or how to do that.
THANK YOU VERY MUCH FOR YOUR TIME AND EFFORT. I really appreciate it, man. Thanks again.