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Product rule calculus help

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100
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y = x ln x

[tex]\frac{dy}{dx} = \frac{1}{x ln x} \cdot 1[/tex]

Is that correct?
 
609
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wrong... use product rule

[tex] \frac{d}{dx} (fg) = \frac{df}{dx}g+f\frac{dg}{dx} [/tex]
 
100
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And how would that look like? Which is f & which one is g? :confused:
 
Hmm i'm not great at this but,

y = x lnx
I was gunna say.. that whenever you take the derivative of logs.. 1/function * derivative of that function
so that'd give you
y = (1/xlnx)(1/x)
but by using the product rule...
y = 1(xlnx) + x(1/x))
Hmm, definately don't listen to me, i don't know WHats going on. Here to learn!
 
100
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SphericalStrife said:
Hmm i'm not great at this but,

y = x lnx
I was gunna say.. that whenever you take the derivative of logs.. 1/function * derivative of that function
so that'd give you
y = (1/xlnx)(1/x)
I was taught that way. How did you get 1/x?
 
hmm I think i might just be being stupid. 1/x is the known derivative of lnx.. but i guess with that other x there.. x lnx, you would need to use product rule like that guy said?
 
100
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Oh I see...
SphericalStrife said:
1/x is the known derivative of lnx
How do you know it?
 
147
1
Use product rule. f(x)=x and g(x)=ln(x)
Now just find derivatives
y = x'*lnx + x*ln(x)'
 
100
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Yeah I got it. Thank you.
 

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