Posted by **Jill** on Saturday, November 13, 2010 at 5:16pm.

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!

- Math--Calculus -
**Jill**, Saturday, November 13, 2010 at 5:18pm
Oh and in addition,

a= 12,

b= 2/15.

Thanks.

- Math--Calculus -
**MathMate**, Saturday, November 13, 2010 at 8:37pm
"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(1-tan((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

- Math--Calculus -
**Jill**, Sunday, November 14, 2010 at 12:54am
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.

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