Hi, one of my problems for technical calculus is:

Find the derivative of Y=ln(3xcosx)

My logarithmic examples have not included trig functions. I am thinking I will have to set this up as product rule somehow and then take the natural log, correct?

y1= (3x) [cosx] + (cos x) [3x]
y1= 3cosx + (-sinx) (3)
y1= 3cosx - 3sinx
if that is correct, how do I take the natural log of trig functions?

Thanks so much. I have tried using my notes and book but none of the examples were set up like this.

If y= \ln(f(x)) then the derivative is:

y'(x) = \frac 1 {f(x)} f'(x)

For this problem you have f(x) = 3x \cos(x). Try it again, and post back with your results.

If y= \ln(f(x)) then the derivative is:

y'(x) = \frac 1 {f(x)} f'(x)

For this problem you have f(x) = 3x \cos(x). Try it again, and post back with your results.


If I am understanding you (which I might not be because I am terrible with math)

y1= 1/ [(3x)(-sinx)]

To get rid of -sin, change to

y1= (1/-3xsinx)
or -(1/3xsinx)

No. f(x) = 3x cos(x), So f'(x) = 3x(-sin(x))+3cos(x).

Let me show you a different but similar problem, so you can get the idea: find the derivative of y= \ln( \tan(x)). This is of the form y= \ln(f(x)) where f(x) =\ tan(x). Recall that the derivative of tan(x) is 1/cos^2(x), so:

y'(x) = \frac 1 {f(x)} f'(x) = \frac 1 {\tan(x)} \ \frac 1 {\cos ^2(x)} = \frac 1 {\sin(x) \cos(x)}


You can use this same technique for your problem.