How to Use Logarithmic Differentiation in Calculus

[A_Munk3y]

Homework Statement

Hi again
Practicing log differentiation now :P

Sorry for all the questions, i have a midterm on thursday and i need to get this stuff by then

1. e^x2ln(2x+1)
2.ln(2x/(2x+1))
3. (3x-2)^x (use logarithmic differentiation even though unnecessary

The Attempt at a Solution

1. e^x2ln(2x+1)
=>e^x2*2x*(2/(2x+1))
is that right?

2. ln(2x/(2x+1))
=>ln(2x)-ln(2x+1)
=>2/(2x)-2(2x+1)
and is that one right? :)

3. (3x-2)^x
f(x)=(3x-2)^x
ln(y)=ln[((3x-2)^x]
=>ln(y)=xln(3x-2)
=>1/y*y'=1*1/(3x-2)*3
=>1/y*y'=3/(3x-2)
=>y'=y(3/(3x-2)
=>y'=(3x-2)^x(3/(3x-2)

 

[Mentallic]

You've forgotten how to use the product rule.

$$y'=u'v+v'u$$

Where u and v are some functions of x.

1) let $$u=e^{x^2}$$ and $$v=ln(2x+1)$$

2) assuming you meant 2/(2x+1) then yes it's right. By the way, it's nice to see that you converted the log(a/b) into log(a)-log(b) first and then took the derivative. This is a much faster way than taking the derivative of the quotient, not to mention less prone to mistake.

3) For xln(3x-2) use the product rule.

 

[A_Munk3y]

wow :( that's embarassing lol

so is it right now?
1. 1. e^x2ln(2x+1)
=>e^x2*2/(2x+1)+ln(2x+1)*e^x2*2x3. (3x-2)^x
f(x)=(3x-2)^x
ln(y)=ln[((3x-2)^x]
=>ln(y)=xln(3x-2)
=>1/y*y'=x*1/(3x-2)*3+ln(3x-2)*1
=>1/y*y'=3/(3x-2)+ln(3x-2)
=>y'=y[(3/(3x-2))+ln(3x-2)]
=>y'=(3x-2)^x[3/(3x-2)+ln(3x-2]
??
hopefully lol, this is so much confusion!

 

[Mentallic]

Yes you got it right

=>1/y*y'=x*1/(3x-2)*3+ln(3x-2)*1
=>1/y*y'=3/(3x-2)+ln(3x-2)

You accidentally got rid of the x between these two lines. 3/(3x-2) should be 3x/(3x-2).

 

[A_Munk3y]

:D
Yup, didn't pay attention to that.
Ok, thank you so much for all the help mate :)
I really appreciate it

 

[Mentallic]

No worries. Good luck!