Calculus
posted by Jane .
Need help to find dy/dx:
1) y = (sin x) ^1/3
answer dy/dx =  1/ 3(sinx) ^4/3
2) y = sqrt 8 + sin(6x)
answer  lost

1) close, you left out the derivative of the base, so
dy/dx =  1/ 3(sinx) ^4/3(cosx)
for 2)
rewrite it as
y = (8+sin(6x))^1/2
dy/dx = (1/2)((8+sin(6x))^1/2)(cos(6x))(6)
= 3cos(6x)((8+sin(6x))^1/2)
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