Question

Can someone please help me with this......
If y=x(lnx)^2 then (dy/dx)=

Answer 1

Product rule. d(uv)=u'v+uv'. Also, don't forget to use the chain rule when calculating the derivative of (ln x)^2.

Answer 2

\[\begin{align*}y&=x(\ln x)^2\\
\frac{dy}{dx}&=\frac{d}{dx}\left[x(\ln x)^2\right]\\
&=\frac{d}{dx}[x]\cdot(\ln x)^2 + x\cdot\frac{d}{dx}\left[(\ln x)^2\right]
\end{align*}\]
Applying power rule and chain rule on the right-most term, you have
\[\frac{d}{dx}\left[(\ln x)^2\right]=2\ln x\cdot \frac{d}{dx}[\ln x]\]

Answer 3

would the answer be (lnx)(2+xlnx)

Answer 4

is that the answer

Answer 5

It's close, but no.

Answer 6

what is wrong

Answer 7

You shouldn't have a power of \(x\) in your final answer. Remember, you have
\[\color{red}{x}\cdot2\ln x\cdot\color{blue}{\frac{d}{dx}\left[\ln x\right]}\]
What can you tell about the highlighted terms?

Answer 8

(lnx)(2+lnx)

Answer 9

Yup, that's it.

Answer 10

ok thanks

Answer 11

You're welcome