Differentiating Cosines with Cosines?

$\begin{aligned} X & = \left[ \frac{d}{d \cos \theta} \sum_{n=1}^{100} \cos n \theta \right]_{\theta=2\pi} \\ & = \left[ \frac{d}{d\color{#3D99F6}{x}} \sum_{n=1}^{100} \color{#3D99F6}{T_n(x)} \right]_{\color{#3D99F6}{x}=1} \quad \small \color{#3D99F6}{\text{where } T_n(x) \text{ is Chebyshev polynomial of the first kind}} \\ & = \left[ \sum_{n=1}^{100} \frac{d}{dx}T_n(x) \right]_{x=1} \\ & = \left[ \sum_{n=1}^{100} n \color{#3D99F6}{U_{n-1} (x)} \right]_{x=1} \quad \small \color{#3D99F6}{\text{where } U_n(x) \text{ is Chebyshev polynomial of the second kind}} \\ & = \left[ \sum_{n=1}^{100} \frac{n \color{#3D99F6}{\sin n \theta}}{\color{#3D99F6}{\sin \theta}} \right]_{\theta \to 2 \pi} \\ & = \left[ \sum_{n=1}^{100} \frac{n^2 \times \frac{\sin n \theta}{n \theta}}{\frac{\sin \theta}{\theta}} \right]_{\theta \to 0} \\ & = \sum_{n=1}^{100} n^2 = \frac{100(101)(201)}{6} = \boxed{338350} \end{aligned}$

More about Chebyshev polynomials

Note: Without use of Chebyshev polynomial.

$\displaystyle\large\left[ \frac { d }{ d\left( \cos { x } \right) } \left(\displaystyle \sum _{ n=1 }^{ 100 }{ \cos { nx } } \right) \right] _{ { x=2\pi } }=\left[ \frac { d\left( \cos { x } \right) }{ d\left( \cos { x } \right) } +\frac { d\left( \cos { 2x } \right) }{ d\left( \cos{ x } \right) } +...+\frac { d(\cos { 100x) } }{ d(\cos { x) } } \right] _{ x=2\pi }$ .

Since the function is same at $x=2\pi$ and $x=0$ , we replace it by $x=0$ .

$\displaystyle \large[\frac{-sinx}{-sinx}+\frac {-2sin2x}{-sinx}+.....+\frac {-100sin100x}{-sinx}]_{x=0}$

Consider the general term as $\frac{-nsin(nx)}{-sinx}$ Multiply by $\frac {nx}{nx}$ and take limit at $x=0$ .

$lim_{x->0} \frac {nsin (nx)}{sinx}×\frac {nx}{nx}$ . (Use $lim_{x->0} \frac {sinx}{x}$ =1), we get each term as $n^2$ .

So answer is $1^2+2^2+....+100^2=\frac {100×101×201}{6}=\boxed {338350}$

$\displaystyle\large\left[ \frac { d }{ d\left( \cos { x } \right) } \left(\displaystyle \sum _{ n=1 }^{ 100 }{ \cos { nx } } \right) \right] _{ { x=2\pi } }=\left[ \frac { d\left( \cos { x } \right) }{ d\left( \cos { x } \right) } +\frac { d\left( \cos { 2x } \right) }{ d\left( \cos{ x } \right) } +...+\frac { d(\cos { 100x) } }{ d(\cos { x) } } \right] _{ x=2\pi }$ .

$\text{The given expression can also be written as}$ $\displaystyle\large\sum _{ n=1 }^{ 100 }{ { T' }_{ n } } (1)$ , $\text{where}$ ${ T }_{ n }$ $\text{represents the}$ $\text{nth}$ Chebyshev Polynomial of the first kind .

$\text{Using the property}$ , $\displaystyle\large{ T' }_{ n }(1)={ n }^{ 2 }$ , $\text{we get}$ $\displaystyle\large\sum_{ n=1 }^{ 100 }{ { { T' }_{ n } } } (1)=\sum _{ n=1 }^{ 100 }{ { n }^{ 2 } } =338350$ .

i do not know what is Chebyshev Polynomial of the first kind. but did using de-moivre's theorms . when you turn them in exponential and turn every term in cos x and then differentiate it ! not stylish :P @Swapnil Das