- #1

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y = x ln x

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

Is that correct?

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

Is that correct?

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- #1

- 100

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y = x ln x

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

Is that correct?

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

Is that correct?

- #2

- 609

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wrong... use product rule

[tex] \frac{d}{dx} (fg) = \frac{df}{dx}g+f\frac{dg}{dx} [/tex]

[tex] \frac{d}{dx} (fg) = \frac{df}{dx}g+f\frac{dg}{dx} [/tex]

- #3

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

- #4

- 8

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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!

- #5

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I was taught that way. How did you get 1/x?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)

- #6

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- #7

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Oh I see...

How do you know it?SphericalStrife said:1/x is the known derivative of lnx

- #8

- 147

- 1

Use product rule. f(x)=x and g(x)=ln(x)

Now just find derivatives

y = x'*lnx + x*ln(x)'

Now just find derivatives

y = x'*lnx + x*ln(x)'

- #9

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Yeah I got it. Thank you.

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