Finding dy/dx for y = x ln x

[footprints]

y = x ln x

$$\frac{dy}{dx} = \frac{1}{x ln x} \cdot 1$$

Is that correct?

 

[vincentchan]

wrong... use product rule

$$\frac{d}{dx} (fg) = \frac{df}{dx}g+f\frac{dg}{dx}$$

 

[footprints]

And how would that look like? Which is f & which one is g?

 

[SphericalStrife]

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, definitely don't listen to me, i don't know WHats going on. Here to learn!

 

[footprints]

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?

 

[SphericalStrife]

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?

 

[footprints]

Oh I see...

1/x is the known derivative of lnx

How do you know it?

 

[Yegor]

Use product rule. f(x)=x and g(x)=ln(x)
Now just find derivatives
y = x'*lnx + x*ln(x)'

 

[footprints]

Yeah I got it. Thank you.