Posted by Jill on .
I'm having a tough time figuring out this problem...
S(x) = bracket (piecewise function)
a + b arcsin*(tan x/tan 66) for 0 ≤ x < 66
24 for 66 ≤ x ≤ 90.
Is the function differentiable? Why or why not?
Could someone please help me? My teacher told me the function was not differentiable, but I need to prove why it is not. Thank you!

MathCalculus 
Jill,
Oh and in addition,
a= 12,
b= 2/15.
Thanks. 
MathCalculus 
MathMate,
"S(x) = bracket (piecewise function)
a + b arcsin*(tan x/tan 66) for 0 ≤ x < 66
24 for 66 ≤ x ≤ 90. Given a=12, b=2/15.
Is the function differentiable? Why or why not?"
I suppose this is the number of hours of sunshine where x is the latitude in degrees during solstice (summer or winter). I have not checked the validity of the equation.
Yes, your teacher is right, the function is not differentiable, but only at x=66.
Recall that the conditions for differentiability are:
1. the function must be continuous. This condition is satisfied throughout the domain 0≤x≤90, notably even at x≥66.
2. The derivative must exist at all points of the domain [0,90]. This is true.
For [0,66),
f(x)=a + b arcsin*(tan x/tan 66), and
f'(x)=(2*sec((π*x)/180)^2)/(15*tan((11*π)/30)*sqrt(1tan((π*x)/180)^2/tan((11*π)/30)^2)) ...(1)
exists.
For [66,90], f'(x)=0.
3. For every point c on the domain,
the derivative from the left must equal the derivative from the right, and equal to the derivative at the point c, or
f'(c)=f'(c)=f'(c+)
This condition is not satisfied at c=66.
To calculate the derivative from the left, we resort to equation (1), which gives the derivative as x>66 to be ∞. As an example, the derivative at 65.999 evaluates to 37.
The derivative at c, and to the right of c is 0.
Therefore the function is not differentiable at x=66.
See:
http://img193.imageshack.us/img193/7569/1289686601.png 
MathCalculus 
Jill,
Ohh, I understand now! Thank you SO much. Your explanation was clear as day, and I cannot tell you just how much I appreciate it.