1. ## derivative ln...

Y=x(lnx)^2

I was thinking you would use product rule, and do this:

y'=x*2(lnx)+(lnx)^2*1

y'=2x(lnx)+(lnx)^2

but the answer in the book, is:

y'=2(lnx)+(lnx)^2

What am I doing wrong?

2. ## Re: derivative ln...

Originally Posted by StudentMCCS
Y=x(lnx)^2

I was thinking you would use product rule, and do this:

y'=x*2(lnx)+(lnx)^2*1

y'=2x(lnx)+(lnx)^2

but the answer in the book, is:

y'=2(lnx)+(lnx)^2

What am I doing wrong?
Did you use the chain rule when finding $\dfrac{d}{dx} (\ln(x))^2$

3. ## Re: derivative ln...

$y=x(\ln{x})^2$ so $\frac{dy}{dx}=(\ln{x})^2+x\frac{d}{dx}(\ln{x})^2$

you made a mistake while solving $x\frac{d}{dx}(\ln{x})^2$.

To find : $x\frac{d}{dx}(\ln{x})^2$ we will take $g = (\ln{x})^2$ and $u = \ln{x}$.

$x\frac{d}{dx}(\ln{x})^2=x\frac{dg}{dx}=x\frac{du}{ dx}\frac{dg}{du}=x.\frac{1}{x}.(2\ln{x})=2\ln{x}$

hence $\frac{dy}{dx}=(\ln{x})^2+2\ln{x}$

do you understand now?