For a), you can't just add up derivatives to make higher order derivatives. As each fourth derivative is the original function again, you just have to take 999 modulo 4 = 249 * 4 + 3 modulo 4 = 3. And the answer is the 3rd derivative.
Question :
(a) let y = cos x, find d^999y/dx^999
(b) let y = x^x^x^x^x^x...., find dy/dy
(c) let y =sqrt(x+sqrt(x+sqrt(x+sqrt(x+......), find dy/dx
my attempt :
(a) y= cos x
y' = -sin x
y'' = -cos x
y'''= sin x
y^(IV) =cos x = y
thus, y^(4n) = y
then 999/4 = 249 bal 3
999 = (249x4)+3
thus, d^999y/dx^999 = y^(999)
= y^(249x4)+3
= y^(996).y^(3)
= cos x sin x
(b) y = x^x^x^x^x^x.... can be expressed as y=x^y
taking log both side log y = y log x....(*)
diff * w.r.t. x :
1/y(dy/dx) = y(1/x) +log x (dy/dx)
(1/y-log x )(dy/dx) =y/x
[(1-y log y)/y] dy/dx = y/x
dy/dx =(y^2)/x(1-y log x)
(c) y =sqrt(x+sqrt(x+sqrt(x+sqrt(x+......) can be expressed as y = sqrt(x+y)
squaring both side
y^2 = x+y ....(*)
diff (*) w.r.t. x :
2y(dy/dx) = 1 +dy/dx
(2y-1) (dy/dx) = 1
dy/dx = 1/(2y-1)
IS MY ANSWER CORRECT, i'm afraid my answer for (a) is wrong... can anyone clariffy ot for me??? THANX...
f(x) = cos x
f'(x) = -sin x
f''(x) = -cos x
f'''(x) = sin x
f''''(x) = cos x
...
There is a cycle with period 4, so if we write the order of the derivative as n, and we write n = 4k + r, with r < 4, then to calculate the nth order derivative it is sufficient to calculate the rth order derivative. Congruences with moduli are a notation used to express such things. (Wikipedia can tell you more if you want to know.)