Calculator Derivative Error at 11: What's Causing the Drastic Difference?

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Hi everyone, I've discovered quite a strange function in my textbook. I compared the calculator's calculated derivatives (Math->nDriv) with my calculus derivative and found that the two are almost 2000 units off! Up until now I've been trying to figure out the problem, but to no avail, I cannot find the solution. Here's the function:

f(x)=Cos2(tanx)

using derivative shortcuts, I came up with

f'(x)=-2cos(tanx)sin(tanx)sec2x

as my derivative function

plugging in 11, I came up with -23921.7403... as the derivative at 11

however, when i compare this with the calculator's nDriv answer, it is drastically different:
nDriv(f(x), x, 11)= -342.6

that's difference in the 2000 units... i mean, i compared the rest of the values at various other points and the most off they have been is around .001; only at 11 does this great gap occur.

Does anyone have any idea what the problem is? Any tips and help will be appreciated. If you have a ti-89 or another version of the calculator, please feel free to try it on your calculator and post if you have the same result as me.

Thanks so much,
Alex
 
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i assume you mean
f(x)=cos^2(tan(x))

differentiating
f'(x)=-2cos(tan(x))sin(tan(x)) \frac{1}{cos^2(x)}

2 things to note
cos(11) = 0.00442
which is a small number, as we are close to zero the value of the function will be diverging quite quickly. I would trust the analytic solution...

the other thing to check is whether you are in degrees or radians
 
that's really weird because I've never seen this happen with any other function, is there an example that you can give where the result will be replicated?
 
Any time you are dividing by a very small number (cos(11)= 0.0044257), round off error becomes important.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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