MHB 242 Derivatives of Logarithmic Functions of y=xlnx-x

[karush]

$\tiny{from\, steward\, v8\, 6.4.2}$
find y'
$\quad y= x\ln{x}-x$
so
$\quad y'=(x\ln{x})'-(x)'$
product rule
$\quad (x\ln{x})'=x\cdot\dfrac{1}{x}+\ln{x}\cdot(1)=1+\ln{x}$
and
$\quad (-x)'=-1$
finally
$\quad \ln{x}+1-1=\ln{x}$

well this is an even # with no book answer but I think it is ok
typos maybe
also guess we could of factored out the x but not sure if this would help

 

[skeeter]

$y = x(\ln{x}-1)$

$y' = x \cdot \dfrac{1}{x} + (\ln{x} - 1) \cdot 1 = 1 + \ln{x} - 1 = \ln{x}$

same same

 

[karush]

thot I would slip in another one under the table ...

$y=\ln{\sqrt{\dfrac{a^2-z^2}{a^2+z^2}}}$ok calculator says y'=0 but not sure what you do with the 2 variables

 

[skeeter]

thot I would slip in another one under the table ...

$y=\ln{\sqrt{\dfrac{a^2-z^2}{a^2+z^2}}}$ok calculator says y'=0 but not sure what you do with the 2 variables

I'm assuming $y$ is a function of $z$ and $a$ is a constant.

using properties of logs ...

$y = \dfrac{1}{2}\left[\ln(a^2-z^2) - \ln(a^2+z^2)\right]$

$\dfrac{dy}{dz} = \dfrac{1}{2}\left(\dfrac{-2z}{a^2-z^2} - \dfrac{2z}{a^2+z^2}\right)$

$\dfrac{dy}{dz} = \dfrac{-z}{a^2-z^2} - \dfrac{z}{a^2+z^2}$

common denominator & combine fractions ...

$\dfrac{dy}{dz} = \dfrac{2a^2 z}{z^4-a^4}$

 

[HallsofIvy]

So y'= 0 at z= 0. Another possibility is that your calculator is assuming that the independent variable is always x and treated both a and z as constants so the whole function is treated as a constant so the derivative is 0.