Help on logarithmic differentiation problem

[crm08]

Homework Statement

Use logarithmic differentiation to find the derivative of the function.

Homework Equations

y = (sin(x))^(ln(x))

The Attempt at a Solution

I guessing the first step is raise both sides to "e", but so far I have only completed problems by taking the natural log of both sides.

(1) e^(y) = e^(sin(x))^(ln(x))

(2) d/dx (e^y) = d/dx (e^(sin(x))^(ln(x)))

(3) Can I start moving exponents in front of "e" to use the product rule, if so, do I bring "sin(x)^(ln(x)) down?, or am I on the completely wrong track?

 

[tiny-tim]

Hi crm08!

y = (sin(x))^(ln(x))
…
(1) e^(y) = e^(sin(x))^(ln(x))

= e^(sin(x)*ln(x))

 

[crm08]

Oh I gotcha, so now use:

d/dx (e^g(x)) = e^g(x) * g'(x), where g(x) = (sin(x))(ln(x)), and use product rule for g'(x)

I think I'm on the right track

 

[wsalem]

use ln not e, since \ln{x^a} =a \ln{x}
So that \ln{y} = \ln( \sin{x}^{\ln{x}} ) = \ln(x) \ln(sin{x})

 

[wsalem]

tiny-tim, I'm confused, why would e^y = e^{(\sin(x)^{\ln(x)})} equals e^{(\sin(x)*\ln(x))}

 

[crm08]

ok I'm getting

(1/y) * dy/dx = (ln(x))*(1/sin(x))*(cos(x)) + (ln(sin(x))*(1/x)

= y*[(ln(x))*(tan(x)) + ((ln(sin(x))) / x)]

am I getting any closer?

 

[Gib Z]

tiny-tim, I'm confused, why would e^y = e^{(\sin(x)^{\ln(x)})} equals e^{(\sin(x)*\ln(x))}

It wouldn't. In general, (a^b)^c = a^(bc), but not a^(b^c). Take an example using a=3, b=2, c=3.

(a^b)^c = 729= a^(bc). a^(b^c) = 3^8, different.

I'm sure tiny-tim just got a bit confused.

As for crm08 - the tan should be a cot, and you should not use an equals signs when you actually multiplied by y, but other than that its looking sweet.

 

[tiny-tim]

oops!

I'm sure tiny-tim just got a bit confused.

mmm … all those ^s …

it'll be a lot easier if everyone uses the X² tag (just above the reply box)

let's see, it should have been …

sinx^lnx = (e^ln(sinx))^lnx = e^ln(sinx)*lnx​

 

[wsalem]

ahhh, so you were referring to y and not e^y that equals e^{\ln(sinx)*\ln{x}}.
That makes sense now!