Question 6 (1 point) Find the derivative of f(x)=lnvert x^3cosxvert -tanx+(x)/(3) tanx+(3)/(x) -tanx+(3)/(x) -tanx-(3)/(x) -tanx-(x)/(3)

## Step 1: Applying the Chain Rule<br />### The derivative of $\ln|u|$ is $\frac{u'}{u}$. Here, $u = x^3 \cos x$. So, $f'(x) = \frac{(x^3 \cos x)'}{x^3 \cos x}$.<br /><br />## Step 2: Finding the derivative of $x^3 \cos x$<br />### Using the product rule, $(x^3 \cos x)' = (x^3)'(\cos x) + (x^3)(\cos x)' = 3x^2 \cos x - x^3 \sin x$.<br /><br />## Step 3: Simplifying the derivative<br />### Substituting the result from Step 2 into Step 1, we get $f'(x) = \frac{3x^2 \cos x - x^3 \sin x}{x^3 \cos x} = \frac{3x^2 \cos x}{x^3 \cos x} - \frac{x^3 \sin x}{x^3 \cos x} = \frac{3}{x} - \tan x$.