logarithmic differentation

[Nidhogg]

how do i find the derivative of this using logarithmic differentation?

y=(ln(sinx))^x

^x is to the power of x

 

[stapel]

Just apply the log-diff process, as outlined in class and in your text:

. . . . .Take the natural log of both sides.

. . . . .Differentiate both sides with respect to "x".

. . . . .Isolate "dy/dx" on one side.

. . . . .Replace every instance of "y" on the other
. . . . .side with the original expression for "y"
. . . . .(from before having taken the logs).

How far have you gotten in this process?

Thank you.

Eliz.

 

[Nidhogg]

lny=xln(ln(sinx))

1/y*y' = xln(ln(sinx))

im not sure what to do then

 

[stapel]

Now you need to differentiate the right-hand side.

Once that is done, multiply through by "y" to get "dy/dx" by itself on the left-hand side.

Then replace the "y" with "(ln(sin(x)))^x" to get the final answer.

Eliz.

 

[Nidhogg]

y=(ln(sinx))^x

1/y*y' = xln(ln(sinx)

1/y*y' = x*(cotx/ln(sinx))+ln(ln(sinx))

y' = (ln(sinx))^x *(x(cotx/ln(sinx))+ln(ln(sinx))

is that right?

 

[soroban]

Hello, Nidhogg!

$\displaystyle y \:=\:\left[\ln(\sin x)\right]^x$

$\displaystyle \ln(y) \:=\:x\cdot\ln\left[\ln(\sin x)\right]$

$\displaystyle \frac{1}{y}\cdot y' \:= \:x\cdot\frac{\cot x}{\ln(\sin x)}\,+\,\ln\left[\ln(\sin x)\right]$

$\displaystyle y' \:= \:\left[\ln(\sin x)\right]^x\cdot\left(\frac{x\cot x}{\ln(sin x)} \,+\,\ln\left[\ln(\sin x)\right]\right)$

Is that right?

Looks good to me . . .Good work!