can some one help me on C3 differentiation?Watch

Thread starter 13 years ago
#1
can someone show me the steps to differentiate [x+ln(2x)]^3. I work it out to be 1.5[1+0.5x]^2 but I think this is wrong. Thanks!
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13 years ago
#2
If I had to guess, I would have said 3[1+0.5x]^2, but I haven't really done much Calculus yet...
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Thread starter 13 years ago
#3
anyone else?
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13 years ago
#4
I havent done any differentiation in quite a while but i wouold have said it was

[(6x+3)/2x][x+ln|2x|]^2
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13 years ago
#5
y = u^3
u = x+ln(2x)

dy/du = 3u^2
du/dx = 1+1/x

dy/du*du/dx = (3u^2)(1+1/x)

3(x+ln(2x)^2(1+1/x)

Hopefully that's correct

d/dx[ln(2x)] = 2*1/2x = 2/2x = 1/x

You can show this by saying

y = lnu
u = 2x

dy/du = 1/u
du/dx = 2

dy/dx = dy/du*du/dx = 1/2x*2

Which is 1/x
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13 years ago
#6
(Original post by lucaz)
can someone show me the steps to differentiate [x+ln(2x)]^3. I work it out to be 1.5[1+0.5x]^2 but I think this is wrong. Thanks!
let (x+ln2x)=u and (x+ln2x)3 =v

du/dx= 1+(1/x)
dv/dx= 3(x+ln2x)2 x (1+(1/x))
=3(x+ln2x)2 (1+(1/x))
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13 years ago
#7
(Original post by SunGod87)
y = u^3
u = x+ln(2x)

dy/du = 3u^2
du/dx = 1+1/x

dy/du*du/dx = (3u^2)(1+1/x)

3(x+ln(2x)^2(1+1/x)

Hopefully that's correct

d/dx[ln(2x)] = 2*1/2x = 2/2x = 1/x

I think :P
I agree. 3[{x+ln(2x)}^2](1+1/x) Thats the same as I got, i just simplified it.
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13 years ago
#8
yay
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