derivative of a function

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...(Speechless)

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.

Quote:

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Originally Posted by brouwer

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.

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can you explain why we only considered the 3rd derivative as the answer? still blur... what is 'modulo' actually? (Worried)

Quote:

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Originally Posted by bobey

can you explain why we only considered the 3rd derivative as the answer? still blur... what is 'modulo' actually? (Worried)

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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.)