calculate the 998th derivative of f(x)=sin(x^2) of f evaluated at 0.

Question

please help, tried about everything I know and I can't get an answer.

Answer 1

You have to find formula for derivative of order n. To do that you find f''(x), f''(x) and son until you see a pattern from which follows a general formula.

You have f(x)=sin(x^2) So,

f'(x)=2x cos(x^2),

f''(x) = {2 cos[x^2] - 4 x^2 Sin[x^2],

f''(x) = -8 x^3 Cos[x^2] - 12 x Sin[x^2],

f^(4)(x) = -48 x^2 Cos[x^2] - 12 Sin[x^2] + 16 x^4 Sin[x^2],

f⁽⁵⁾(x) = -120 x Cos[x^2] + 32 x^5 Cos[x^2] + 160 x^3 Sin[x^2]

and so on.

Now look if you can see a common pattern for the derivative of any order. If yes, write down the general formula. if not compute derivative of higher orders.

OR

After taking derivative of each order you can evaluate its value at 0, that is you'll find

f(x), f;(0), f''(0x) and so on. Try. But at any case you need to take derivatives..