1. Logarithmic Differentiation?

Use logarithmic differentiation to find the derivative of $\displaystyle y=\frac{{x}^{19} (cos(3x)-11)}{2x^3-7}$

I was wondering how to handle the little 3x within cosine, do I chain rule it? If I do I end up with this:

$\displaystyle \frac{dy}{dx}=\frac{{x}^{19}(cos(3x)-11)}{2x^3-7} * (\frac{19}{x} - \frac{sin(3x)}{xcos(3x)-11}-\frac{6x^2}{2x^3-7})$

Is there anything wrong?

2. 1) Yes, use the chain rule. What other plan have you?
2) I'm having trouble seeing the logarithm. Perhaps doing what it asks will greatly simplify your efforts. This will give you even more chain rule practice.

3. well I multiplied at the section $\displaystyle \frac{sin(3x)}{cos(3x)-11}$ with $\displaystyle \frac{1}{3x}$ and 3. The 3 cancels leaving the x with the cosine, no? Or is it the 3 that's supposed to stay?

4. oh no wait I made a mistake, I think this is it:

$\displaystyle \frac{dy}{dx}=\frac{{x}^{19}(cos(3x)-11)}{2x^3-7} * (\frac{19}{x} - \frac{3}{cos(3x)-11}-\frac{6x^2}{2x^3-7})$

5. I think they want you to take the log of both sides then take the derivative. That would simplify things a bit:

$\displaystyle ln(y) = 19ln(x) + ln(cos(3x) - 11) - ln(2x^3-7)$

Then differentiate w.r.t. x.

6. right so $\displaystyle 19\frac{1}{x}+-sin(3x)(3)\frac{1}{cos(3x)-11}-6x^2\frac{1}{2x^3-7}$

7. Differentiate the left-hand side of the equation as well. You should get $\displaystyle \frac{\frac{dy}{dx}}{y}$ equals what you computed. Then solve for [tex] \frac{dy}{dx} [\tex].

8. right so I just bring back the original function and multiply it to what I have now, right?