derivative ln...

**StudentMCCS** · October 16th 2011, 02:06 PM

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?

**e^(i*pi)** · October 16th 2011, 02:13 PM

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 $\dfrac{d}{dx} (\ln(x))^2$

**sbhatnagar** · October 16th 2011, 11:55 PM

$y=x(\ln{x})^2$ 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?

Thread: derivative ln...

LinkBack

Thread Tools

Display

derivative ln...

Re: derivative ln...

Re: derivative ln...

Similar Math Help Forum Discussions

contuous weak derivative $\Rightarrow$ classic derivative ?

Expanding a time derivative in terms of other variables /// Material derivative

[SOLVED] Definition of Derivative/Alt. form of the derivative

derivative of 1/[sqrt[x]: proving with derivative definition

Derivative of definate intagral / second derivative? idk.

Search Tags