Find the 57th derivative of the function of f(x)=sin(9u)?

57 mod 4 == 1, so the 57th derivative will be...

f_57 [sin(9u)] = 9^57*cos(9u).

We can take advantage of the cyclic nature of the trig functions.

sin --> cos --> -sin --> -cos --> sin, etc.

We also know that the derivative of sin(kx) = k*cos(kx), so the coefficient keeps getting multiplied by itself.

sin cycles it's derivatives in 4's...

f' = cos

f'' = -sin

f''' = -cos

f^(4) = sin

So, let's divide 57 / 4 to figure out where we'll be.

57 / 4 = 14.25

That tells you that we've cycled through the these 14 times, and just need to take the first derivative.

cos(9u)

Now, using the chain rule, 57 times, we'll have

9^(57) out front, so your answer will be

9^(57) cos(9u)