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anonymous
 5 years ago
find the derivative of:
y= ln[(3x1)^3 (4x+2)^5)
anonymous
 5 years ago
find the derivative of: y= ln[(3x1)^3 (4x+2)^5)

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = ln(u^3 v^5) u = 3x1 ; v=4x+2 or y = ln(uv) ; u = (3x1)^3 v=(4x+2)^5

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0eiher way uv is on the bottom ans the top works out to be u'v' right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i have to use the power rule?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0for the top yes; Dx(uv)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank yos so much. i just get confused when choosing which rule to use. =D

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0uv' + u'v the top is: [(3x1)^3] [20(4x+2)^4] + [9(3x1)^2][(4x+2)^5] might wanna clean that up some tho...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im going to work it out using the formula n then im going to check it up with your answer. thank you so much ^.^
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