applications of calculus to the physical world (1 Viewer)

[chen yi]

I am stuck on this question. It is related to applications of calculus to the physical world somehow. Can anyone show me the way?
Prove d/dx (x lnx) = 1 + lnx.

 

[hyparzero]

uh... it just wants you to differentiate

xlnx with respect to x

Just use the product rule:
Let u = x v= lnx
u' = 1 and v' = 1/x

Therefore d/dx (x lnx) = uv' + vu' = x(1/x) + 1(lnx)

= 1 + lnx

 

[kimii]

you know how u differentiate it...
can u also use the product rule?
which would be
y = x lnx
y'= lnx + x . 1/x
y'= 1 + lnx

 

[Riviet]

you know how u differentiate it...
can u also use the product rule?
which would be
y = x lnx
y'= lnx + x . 1/x
y'= 1 + lnx

That's basically the same as what hyparzero had, it's just that he did more working.

 

[vafa]

an alternative method:

let y=xlnx
take log of both sides

lny=lnx+ln(lnx)
1/y.dy/dx=1/x+(1/x)/lnx

dy/dx=xlnx(1/x+1/xlnx)
dy/dx=xlnx((lnx+1)/xlnx
dy/dx=lnx+1

the second alternative method:

let y= (e^lnx)(e^ln(lnx))

y= e^(lnx+ln(lnx))
dy/dx=(1/x+(1/x)/lnx).xlnx

dy/dx=((lnx+1)/xlnx).xlnx
dy/dx=lnx+1