Find the first five nonzero term using the Maclaurin series

The function is cos(√x) and the problem is that every derivative of cos(√x) is valued as undefined when I use f^n(0). What am I supposed to do?

"every derivative of cos(√x) is valued as undefined"

That's not true at all.

* f(x) = cos(√x) so f(0) = 1
* f'(x) = -sin(√x)/(2√x). What you do is take the limit as x->0. Top and bottom are 0 so you can use L'Hopitals rule (ie, take derivative of the top and take derivative of the bottom).

So lim f'(x) = lim (-cos(√x)/2√x) / (1/√x) = lim (-cos√x)/2 = -1/2.

This starts getting messy. What I would do is use implicit differentiation where we rewrite f(x) = cos(√x) as f(x^2) = cos(x).

* f(x^2) = cos(x),
so f(0) = 1

* 2x f'(x^2) = -sin(x)
=> f'(x^2) = -sin(x)/2x.

As x->0,
f'(0) = lim f'(x^2)
= lim -sin(x)/2x
= lim -cos(x)/2 (Used L'Hopital's rule)
= -1/2

* 2 f'(x^2) + 4x^2 f''(x^2) = -cos(x)
=> f''(x^2) = (-cos(x) -2f'(x^2))/4x^2 = (-cos(x) + sin(x)/x)/4x^2 = (sin(x) - xcos(x))/4x^3.

Again as x->0 top and bottom are zero so you an use L'Hopitals's rule again:

f''(0) = lim f''(x^2)
= lim (sin(x) - xcos(x))/4x^3
= lim (cos(x) - cos(x) + xsin(x))/12x^2 (Used L'Hopital)
= lim sin(x)/12x
= 1/12. (Used L'Hopital)

* 4x f''(x^2) + 8x f''(x^2) + 8x^3 f'''(x^2) = sin(x)
=> f'''(x^2) = (sin(x) - 12x f''(x^2))/8x^3 = (sin(x) - 3(sin(x) - xcos(x))/x^2)/8x^3 = (x^2 sin(x) - 3sin(x) + 3xcos(x))/8x^5.

As x->0,

f'''(0) = lim f'''(x^2)
= lim (2x sin(x) + x^2 cos(x) -3cos(x) + 3 cos(x) -3xsin(x))/40x^4
= lim (x cos(x) - sin(x))/40x^3
= lim ( cos(x) -xsin(x) - cos(x))/120x^2 (Used L'Hopitals)
= lim ( -sin(x)/120x)
= -1/120 (Used L'Hopitals)

You can do the last one.