Derivative and domain of f(x)=xln(sec(x))

This is a composition of functions, so we have to use the chain rule.

Notice that we have a product between x and ln(sec(x)). We also have a logarithm to differentiate and the inside of that is a trigonometric function.

So we use the product rule first - differentiate only x and sum with differentiating only ln(sec(x)). After differentiate ln(sec(x)), we use chain rule.

Your derivative should then be:

f'(x) = ln(sec(x)) + x(1/sec(x))(sec(x)tan(x)) = ln(sec(x)) + xtan(x)

To find the domain, we simply find where the denominator is non-zero. If we can find where it is zero, the rest is non-zero.

We know that sec(x) = 1/cos(x), and that tan(x) = sin(x)/cos(x). We set the denominators to 0. The denominators are 0 when x = pi/2 + 2kpi, and x = -pi/2 + 2kpi (check the unit circle to confirm this) where k is an integer.

Therefore, the domain is all real numbers excluding the one mentioned above (x not equal to ...)

I hope this helped!

f(x) = x·ln(sec(x))

f(x) is the product of x & ln(sec(x)).
x can be anything without making f(x) undefined. So, no restrictions there.
However, ln(sec(x)) is restricted in that the argument of a logarithm must be positive. So, sec(x) must be positive which requires  sec(x) > 0  ⇒  1/cos(x) > 0  ⇒  cos(x) > 0. Since cosine is positive only in the 1st and 4th Quadrants, we can place the angle x in the 1st and 4th Quadrants so that  -π/2 < x < π/2  ⇒  x ∈ (-2π , 2π)  and in general 
Domain = x ∈ (-2π+2πk , 2π+2πk)  for all integers k.

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Note ... Product Rule
 d uv
—— = u·v' + v·u'
 dx


d x·ln(sec x)
————— = x•(1/sec x)·sec x·tan x  +  ln(sec x)•1  =  x·tan x  +  ln(sec x)
       dx




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