Question

find the derivative.
y=x(sin(lnx)+cos(lnx))

y' = xsin(lnx) + xcos(lnx)

y' = xsin(lnx) + x ( 1 / x ) cos(lnx) + xcos(lnx) + x (1 / x ) ( - sin(lnx))

This is my answer.
y' = xsin(lnx) + cos(lnx) + xcos(lnx) -sin(lnx)

The book has an answer of 2cos(lnx) Could someone show me where I'm messing up?

Answer 1

This is what we're differentiating?
\[\Large\rm y=x\left[\sin(\ln x)+\cos(\ln x)\right]\]

Answer 2

the second line is distributing the x in, right? so that the left hand side is y still, right?

Answer 3

So product rule gives us:\[\Large\rm y'=\left[\sin(\ln x)+\cos(\ln x)\right]+x\color{royalblue}{\left[\sin(\ln x)+\cos(\ln x)\right]'}\]It looks like maybe you took a different approach, distributing the x first?
I'm not quite sure..

Answer 4

Differentiating the blue part,\[\Large\rm \color{royalblue}{\cos(\ln x)\frac{1}{x}-\sin(\ln x)\frac{1}{x}}\]

Answer 5

The x in front of the blue stuff cancels with the 1/x's.
So we're left with:\[\Large\rm y'=\sin(\ln x)+\cos(\ln x)+\color{royalblue}{\cos(\ln x)-\sin(\ln x)}\]

Answer 6

Oh boy, the website is tweaking out... :(

Answer 7

the third line is starting take derivative, right?
\[y' = \color{red}{x'}sin(lnx) + x\color{red}{(sin(lnx))'}+ \color{red}{x'}cos(lnx) + x\color{red}{(cos(lnx))'}\]
\[= sin(lnx) + x\color{red}{(sin(lnx))'}+ cos(lnx) + x\color{red}{(cos(lnx))'}\]
\[=sin(lnx) + xcos(ln x)\color{red}{(lnx))'}+ cos(lnx) - xsin(ln x)\color{red}{(lnx)'}\]
\[= sin(ln x) +\dfrac{xcos(ln x)}{x}+cos (lnx )-\dfrac{xsin(lnx)}{x}\]
\[= sin (ln x) + cos (ln x) + cos (ln x) - sin(lnx)\]
= 2 cos (ln x)

Answer 8

Hi Zepdrix and Loser66. Thanks for your help. I figured it out while the website was down. I wasn't deriving the x properly and that's why I was getting it wrong. I appreciate your help though!

Answer 9

np

Answer 10

I suppose using the Definition of Derivative isn't an option.