The answer to this says that ----> x tan x + ln(sec x) with domain equal
to the union of all intervals of the form (-pi/2 + 2pik , pi/2 + 2pik), where k is any integer. So I'm looking for the work to get to this, I'm really confused because an example of this type of question isn't on the lecture! Thank you in advance!!
to the union of all intervals of the form (-pi/2 + 2pik , pi/2 + 2pik), where k is any integer. So I'm looking for the work to get to this, I'm really confused because an example of this type of question isn't on the lecture! Thank you in advance!!
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So to derive this it requires the product rule:
(fg') = f'g + gf'
so f'(x) = ∂/∂x(x)*ln(sec(x)) + ∂/∂x(ln(sec(x))*x
f'(x) = ln(sec(x)) + sin(x)/cos²(x)/sec(x)*x
The sin(x)/cos²(x) comes from the fact that the derivative of ln(f(x)) = f'(x)/f(x), and so the derivative of sec(x) = 1/cos(x) by the reciprocal rule = - (-sin(x))/cos²(x) = sin(x)/cos²(x).
f'(x) = ln(sec(x)) + sin(x)cos(x)/cos²(x)*x
f'(x) = ln(sec(x)) + sin(x)/cos(x)*x
f'(x) = ln(sec(x)) + xtan(x)
so that's where the derivative comes from
The domain is slightly trickier: ln is a unique function such that ln(f(x)) is defined iff and only if f(x) > 0 . However, we know that sec(x) has a range that is R \ (-1, 1) so we must restrict it in such a way such that sec(x) is never negative. If you sketch sec(x), you will see that it has it's first curve above the x-axis between (-π/2, π/2). From (π/2, 3π/2,) it is negative, but from (3π/2, 5π/2) and from (5π/2, 7π/2) it is positive again. So what you can think of this is that if we were to take the portion of the graph of sec(x) that is positive, then it has periodicity 2π - that is it occurs every 2π times. This is why the answer stipulate the domain to be (-π/2 + 2πk, π/2 + 2πk), since they take the first restricted domain where sec(x) > 0, and then they add a period of 2πk to find the next time it is greater than zero.
the key thing to note here is that the natural logarithm (or any logarithm really) function is defined only for values greater than zero, so you must restrict your sec(x) graph to be greater than zero for f(x) to be defined.
(fg') = f'g + gf'
so f'(x) = ∂/∂x(x)*ln(sec(x)) + ∂/∂x(ln(sec(x))*x
f'(x) = ln(sec(x)) + sin(x)/cos²(x)/sec(x)*x
The sin(x)/cos²(x) comes from the fact that the derivative of ln(f(x)) = f'(x)/f(x), and so the derivative of sec(x) = 1/cos(x) by the reciprocal rule = - (-sin(x))/cos²(x) = sin(x)/cos²(x).
f'(x) = ln(sec(x)) + sin(x)cos(x)/cos²(x)*x
f'(x) = ln(sec(x)) + sin(x)/cos(x)*x
f'(x) = ln(sec(x)) + xtan(x)
so that's where the derivative comes from
The domain is slightly trickier: ln is a unique function such that ln(f(x)) is defined iff and only if f(x) > 0 . However, we know that sec(x) has a range that is R \ (-1, 1) so we must restrict it in such a way such that sec(x) is never negative. If you sketch sec(x), you will see that it has it's first curve above the x-axis between (-π/2, π/2). From (π/2, 3π/2,) it is negative, but from (3π/2, 5π/2) and from (5π/2, 7π/2) it is positive again. So what you can think of this is that if we were to take the portion of the graph of sec(x) that is positive, then it has periodicity 2π - that is it occurs every 2π times. This is why the answer stipulate the domain to be (-π/2 + 2πk, π/2 + 2πk), since they take the first restricted domain where sec(x) > 0, and then they add a period of 2πk to find the next time it is greater than zero.
the key thing to note here is that the natural logarithm (or any logarithm really) function is defined only for values greater than zero, so you must restrict your sec(x) graph to be greater than zero for f(x) to be defined.