Logarithmic Differentiation - Please Help

[Naeema F]

What would be the step by step solution to the function? Help would be highly appreciated ?

 

[Dr.Peterson]

Take the log of both sides and differentiate.

Then show us your work.

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[Naeema F]

Hello! Thank you for your reply.
This is what I managed to work out.

ln y = ln e^x + ln cos x - ln arctanx - 5ln 4ln(3x^4)

y'/y = 1/e^x + sinx/cosx - 1/(1+x^2) - 5 ( 12x^3/ ln 4ln 3x^4))


y' = [ 1/e^x + sinx/cosx - 1/(1+x^2) - 5 ( 12x^3/ ln 4ln 3x^4))] y

where y is the given function.

 

[Deleted member 4993]

Hello! Thank you for your reply.
This is what I managed to work out.

ln y = ln e^x + ln cos x - ln arctanx - 5ln 4ln(3x^4)

y'/y = 1/e^x + sinx/cosx - 1/(1+x^2) - 5 ( 12x^3/ ln 4ln 3x^4))


y' = [ 1/e^x + sinx/cosx - 1/(1+x^2) - 5 ( 12x^3/ ln 4ln 3x^4))] y

where y is the given function.

ln y = ln e^x + ln cos x - ln arctanx - 5ln 4ln(3x^4)

This does not look correct (I have not worked it out with pencil and paper).

Next step shown does not follow from the step-above.

use

ln(e^x) = x and

d/dx (cos(x)) = - sin(x)

What does "5ln 4ln(3x^4)" mean? Please simplify!

Please show intermediate steps.

 

[Naeema F]

Thank you for your response. I have no idea how to go about solving this question but I will try.

 

[Deleted member 4993]

Thank you for your response. I have no idea how to go about solving this question but I will try.

As suggested in response#2, you have to calculate "log" of the given function - correctly - first.

You have learned that in pre-calculus.

 

[Dr.Peterson]

Hello! Thank you for your reply.
This is what I managed to work out.

ln y = ln e^x + ln cos x - ln arctanx - 5ln 4ln(3x^4)

y'/y = 1/e^x + sinx/cosx - 1/(1+x^2) - 5 ( 12x^3/ ln 4ln 3x^4))


y' = [ 1/e^x + sinx/cosx - 1/(1+x^2) - 5 ( 12x^3/ ln 4ln 3x^4))] y

where y is the given function.

First line: You seem to have dropped a couple parts (e.g. the exponent of 4 on the whole thing, and the x^2). There is also more simplification you can do.

This is a complicated expression, so take it slowly and check every detail before moving on to differentiation.

Second line: Again, check each part carefully.

Thank you for your response. I have no idea how to go about solving this question but I will try.

Yes, you have plenty of ideas. You just aren't being careful. Be patient with yourself.

 

[Naeema F]

Where do I go from here? Is my solution correct so far?

 

[Dr.Peterson]

The first three lines are good; you are rightly taking it more slowly. But in the last line you have prematurely started differentiating arguments, which is utterly wrong.

Drop that line, and continue expanding the log just as you did originally! Use the quotient property, the product property, and so on, expanding step by step, and when it is fully expanded, then take the derivative of each side.

Incidentally, it helps to keep in mind the reason for what you are doing: taking the log and expanding fully changes products and quotients to sums and differences, which are far easier to differentiate. So take it as far as you can, to save yourself as much work as possible.