Derivative of a Natural Logarithm

[imwishing73]

I just want to make sure I'm doing this correctly...

Find the derivative of ln(xlnx)

(1/xlnx) (1+lnx)<---product rule

=(1+lnx)/(xlnx)

Yes? Not worried about simplification (although I don't think anything needs to be simplified)

Thank you!

 

[Deleted member 4993]

I just want to make sure I'm doing this correctly...

Find the derivative of ln(xlnx)

1/(xlnx) (1+lnx)<---product rule

=(1+lnx)/(xlnx)

Yes? Not worried about simplification (although I don't think anything needs to be simplified)

Thank you!

Looks good to me - except for the grouping symbol as shown.

 

[JeffM]

I just want to make sure I'm doing this correctly...

Find the derivative of ln(xlnx)

(1/xlnx) (1+lnx)<---product rule

You got the right answer, but you used both the product rule and the chain rule.

$\displaystyle u = x * ln(x) \implies \dfrac{du}{dx} = \left\{x * \dfrac{1}{x}\right\} + \{ln(x) * 1\} = 1 + ln(x).$ Used the product rule here.

$\displaystyle y = ln\{x * ln(x)\} = ln(u) \implies \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = \dfrac{1}{u} * \{1 + ln(x)\} = \dfrac{1 + ln(x)}{x * ln(x)}.$ Used chain rule here. Well done.

=(1+lnx)/(xlnx)

Yes? Not worried about simplification (although I don't think anything needs to be simplified)

Thank you!

.