solving a ln equation

[thelazyman]

Hi, can someone please tell me for the derivative of ln(lnx^4) for me, have an idea how to do it but a little unsure.

 

[stapel]

Hi, can someone please tell me for the derivative of ln(lnx^4) for me, have an idea how to do it but a little unsure.

You ask for someone to tell you, regarding the derivative of y = ln(ln(x⁴)), ... what? What do you need with respect to the derivative?

Your subject line refers to solving a log equation. What equation do you need to solve?

Please reply with clarification, including the full and exact (word-for-word) text of the exercise, the complete instructions, and a clear listing of everything you have tried so far.

Thank you.

Eliz.

 

[thelazyman]

The question asks:

Find dy/dx of ln(ln x^4)

Thank you

 

[stapel]

Hint: Apply the Chain Rule.

If you get stuck, please reply showing what you have tried. Thank you.

Eliz.

 

[thelazyman]

can you not use the chain rule? I tried deriving it but got a messed up answer. Can you show me the first step to the chain rule, like what would y = and what would u equal

 

[thelazyman]

Got an answer, not sure if correct.

Ln(ln x^4)


= 1
------------- d (lnx^4)
------
(ln x^4) dx

From that I get

4x^3
--------
x^4
-------
ln x^4

From that I simplify and get an answer of

4
---------

x(ln x^4)

 

[galactus]

Stapel's use of the chain rule isn't in question. The main question is, "Can

you use the chain rule?".

We'll step through. Okey-doke?.


\(\displaystyle \L\\\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\)

\(\displaystyle \L\\y=ln(u)\)

\(\displaystyle \L\\\frac{d}{dx}[ln(u)]=\frac{d}{du}[ln(u)]=\frac{1}{u}=\frac{dy}{du}\)

\(\displaystyle \L\\u=ln(x^{4})\)

\(\displaystyle \L\\\frac{d}{dx}[ln(x^{4})]=\frac{d}{dx}[4ln(x)]=\frac{4}{x}=\frac{du}{dx}\)

Now, put it together. See how that works?. Try applying it to another problem.

EDIT: looks good. That's correct. See?. You got it.

 

[thelazyman]

I got the right answer the way I got it to?

 

[skeeter]

y = ln[ln(x⁴)]

y = ln[4ln(x)]

y = ln(4) + ln[ln(x)]

dy/dx = (1/x)/ln(x) = 1/[xln(x)]