calculus
posted by corinne .
find dy/dx
1. y=(lnx)^3
2. y=x^4lnx
3. y=sin^3(lnx)

1. use chain rule
u = lnx
y = u^3
dy/dx = 3u^2 du/dx = 3(lnx)^2/x
2. use product rule
f=x^4
g=lnx
y = fg
y' = f'g + fg' = 4x^3 lnx + x^4/x = x^3(4lnx + 1)
3. use chain rule
u = lnx
v = sin(u)
y = v^3
dy/dx = dy/dv dv/du du/dx
= 3v^2 * cos(u) * 1/x
= 3sin^2(lnx) * cos(lnx) * 1/x
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