Logarithmic differentiation to find the derivative...

[flaren5]

Hi all,

I have worked through the following question, and I was wondering is anyone would be able to tell me if I'm on the right track, and got the correct answer.

Question:
Use logarithmic differentiation to find the derivative of:

y = 2^(3^(x^(2)))

the way that i solved it is:

lny = ln(2^(3^(x^(2)))
lny =3^(x^(2))ln2
ln(lny) = ln(3^(x^(2))ln2
= ln3(x^(2)) + ln(ln2)
= x^2(ln3) + ln(ln2) --------> is this the correct result?

I woud be very grateful if anyone could tell me if I'm solving this correctly.
Thank you!

 

[stapel]

That's what I get. Now differentiate.

 

[srmichael]

Hi all,

I have worked through the following question, and I was wondering is anyone would be able to tell me if I'm on the right track, and got the correct answer.

Question:
Use logarithmic differentiation to find the derivative of:

y = 2^(3^(x^(2)))

the way that i solved it is:

lny = ln(2^(3^(x^(2)))
lny =3^(x^(2))ln2
ln(lny) = ln(3^(x^(2))ln2
= ln3(x^(2)) + ln(ln2)
= x^2(ln3) + ln(ln2) --------> is this the correct result?

I woud be very grateful if anyone could tell me if I'm solving this correctly.
Thank you!

So far so good!

 

[flaren5]

Thank you, I'm glad to see that I'm on the right track...but when you say, so far so good, and now just differentiate, I thought that I had completed the final anwser. Are you saying that I need to continue, because I have not completed the differentiation?

 

[DrPhil]

Thank you, I'm glad to see that I'm on the right track...but when you say, so far so good, and now just differentiate, I thought that I had completed the final anwser. Are you saying that I need to continue, because I have not completed the differentiation?

What you have done so far is take logarithms and manipulate the equation into a (relatively) simple form. The next phase is to take the derivative (d/dx) of both sides. Then the final steps will be to write the derivative in the form dy/dx = . . .

 

[flaren5]

What you have done so far is take logarithms and manipulate the equation into a (relatively) simple form. The next phase is to take the derivative (d/dx) of both sides. Then the final steps will be to write the derivative in the form dy/dx = . . .

Ok, thank you...so I tried to find the derivative of both sides and this is what I got...

dy/dx = (3^x^2)(X^2)ln3 + xln(ln2) ???

I'm having a hard time deciphering the derivative for the final equation. Does what I have look correct?
Again, thank you for taking the time to repsond...

 

[srmichael]

Ok, thank you...so I tried to find the derivative of both sides and this is what I got...

dy/dx = (3^x^2)(X^2)ln3 + xln(ln2) ???

I'm having a hard time deciphering the derivative for the final equation. Does what I have look correct?
Again, thank you for taking the time to repsond...

Not quite. Yes, the derviative becomes a little messy. Try not to combine steps and you should get the following.

\(\displaystyle ln[ln(y)]=x^2ln3+ln(ln2)\)

\(\displaystyle \frac{d}{dx}ln[ln(y)]=\frac{d}{dx}[x^2ln3+ln(ln2)]\)

\(\displaystyle \frac{1}{lny}\frac{d}{dx}ln(y)=2xln3+0\)

\(\displaystyle \frac{1}{lny}\frac{\frac{dy}{dx}}{y}=2xln3\)

\(\displaystyle \frac{\frac{dy}{dx}}{y}=lny \cdot (2xln3)\)

\(\displaystyle \frac{dy}{dx}=y \cdot [lny \cdot (2xln3)]\)

\(\displaystyle \frac{dy}{dx}=2^{3^{x^{2}}} \cdot 3^{x^{2}}ln2 \cdot (2xln3)\)