Expert: Paul Klarreich - 3/7/2007

Question
I'm sorry, but I have one more question if you have time. I have y=2^xlnx. How do you take the ln of a ln?  If I go ln y = ln (2^xlnx)  and then  I have ln y = xln2 ?  What do I do with the ln of the ln x?  Thanks.
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The text above is a follow-up to ...

-----Question-----
My problem is  g(x)= 4(7^x)  I know that we put "ln" on both sides so we have  ln g(x)= ln 4(7^x). On the right side, when we bring down the x so we have xln7 I am not sure where the 4 goes, what to do with it. I can work the rest of the problem except for that.  Thank you very much.
-----Answer-----
Questioner:   Colleen
Category:  Calculus
 
Subject:  Derivatives of Exponential Functions
Question:  My problem is  g(x)= 4(7^x)  I know that we put "ln" on both sides so we have  ln g(x)= ln 4(7^x). On the right side, when we bring down the x so we have xln7 I am not sure where the 4 goes, what to do with it. I can work the rest of the problem except for that.  Thank you very much.

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Welcome back, Colleen,
Nice to hear from you again.

Here is how you handle this.  Write:

y = 4(7^x)

The basic 'way to handle' the factor of 4 is to fugeddaboutit.  It's just a constant factor of the function, and constant factors get carried through.  So we will just temporarily fugeddaboutit.

y = 7^x

Now it should be easy.

ln y = ln (7^x) = x ln 7
1  dy
--- -- = ln 7
y  dx

the ln 7 is another constant factor, remember.

dy
-- = y ln 7 = 7^x ln 7
dx

OK, you want your 4 back, right.  Your final answer is:

dy
-- = 4 ln 7 7^x
dx
That's  4 * ln 7 * 7^x


Answer
Questioner:   Colleen
Category:  Calculus
 
Subject:  Derivatives of Exponential Functions
Question:  I'm sorry, but I have one more question if you have time. I have y=2^xlnx. How do you take the ln of a ln?  If I go ln y = ln (2^xlnx)  and then  I have ln y = xln2 ?  What do I do with the ln of the ln x?  Thanks.
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Hi again, Colleen,

At the risk of sounding flippant, you take the ln of an ln by pressing the Ln button on the calculator twice.

BUT I think you are referring to a derivative of that.  It just a matter of the Chain Rule again.

y = 2^x ln x

Frankly, since we know from other examples that the derivative of f(x) = a^x  is f' = a^x ln a

we could just use the product rule:

y' = (2^x)(1/x) + (ln x)(2^x ln 2)

y' = (2^x)[1/x  + ln 2 ln x]

But let's try it the old way.  [Or is it the new way?]

Take ln of both sides:
ln y = ln(2^x ln x)

Apply log properties with a vengeance:

ln y = ln(2^x) + ln ln x

ln y = x ln 2 + ln ln x

Now differentiate  ln ln x, which means  ln(ln(x))  if we write it carefully:

********* USUAL WARNING ABOUT COURIER FONT. ********

Use the Chain Rule:

If  r = ln(ln x),  let u = ln x,  du/dx = 1/x
Then r = ln u,  and dr/du = 1/u
dr   dr du    1    1      1
-- = -- -- = ---- --- = ------
dx   du dx   ln x  x    x ln x

Ready to go:

ln y = x ln 2 + ln ln x   [repeated]

1  dy          1
--- -- = ln 2 + ------
y  dx          x ln x

1  dy   x ln x ln 2 + 1
--- -- = ----------------
y  dx      x ln x


dy   (x ln x ln 2 + 1) y
-- = -------------------
dx      x ln x

dy   (x ln x ln 2 + 1) 2^x ln x
-- = --------------------------
dx      x ln x

dy   (x ln x ln 2 + 1) 2^x
-- = --------------------------
dx          x

dy   
-- = (ln x ln 2 + 1/x) 2^x
dx          

which, I believe, is the same answer. [Always a good way to check your differentiation exercise -- do the problem two ways.]