Derivate Next Year

We will split the function into two parts, the $cos(x)$ and $x^{2015}$

a. Let $f(x) = cos(x)$ , then, $f_1(x) = -sin(x); f_2(x) = -cos(x), f_3(x) = sin(x); f_4(x) = cos(x) = f(x)$ From here you will see a bit of recurrent relation with four, thus, to find which product of derivative it will fall under, we just use the n (from n-th) derivative and find the reminder. Here, the reminder of 2016 is 0, thus $f_{2016}(x) = cos(x); f_{2016}(0) = cos(0) = 1$

b. The full solution will not provided here, but please take reference to the question I wrote some time ago. In short, for $x^n$ , the $f_{n}(x) = n!$ , thus $f_{n+1}(x) = 0$

Therefore $f_{2016}(0) = cos(0)+0 = 1+0 = 1$

We will solve the above function in two parts First after every 4th derivative cos x will again become cos x So after 2016 derivative it will again be cos x Second fiction will become as 2016 2015 2014 ........... 3 2 1*0 (x^-1) Which will be zero Therefore f2016(x) =cos 0 + 0 = 1