We Know D^n {cos 2x} = 2^n cos (npi/2 + 2x)
Here n = 50
y ^50 [50 th derivative of y] = 2^50 cos (50 pi/2 + 2x)
= 2^50 cos (25 pi + 2x)
= – 2^50 cos 2x
This was an answer someone posted previously… I don’t understand how they got there… also in general… is there a rule to follow to find a very high derivative for any function.
Thank you!
4 Answers

There are various ways to think about it. Here’s one.
y = cos(2x)
dy/dx = 2sin(2x)
Note to get the derivative we have multiplied the original function by 2 and changed cos to sin.
dy/dx = 2sin(2x) (from previous step)
d²y/dx² = 4cos(2x)
Note to get the new derivative we have multiplied the previous dervative by 2 and changed sin to cos.
d³y/dx³ = 8sin(2x)
Note to get the new derivative we have multiplied the previous derivative by 2 and changed cos to sin.
d⁴y/dx ⁴ = 16cos(2x)
Note to get the new derivative we have multiplied the previous derivative by 2 and changed sin to cos.
After 4 differentiations we are back to (2⁴ = 16 times) the original function. The steps involved were:
A. Multiply by 2. Change cos to sin.
B. Multiply by 2. Change sin to cos.
C. Multiply by 2. Change cos to sin.
D. Multiply by 2. Change sin to cos.
If we differentiate 48 times we will go through 12 lots AD, so we know
d⁴⁸y/dx⁴⁸ = (2⁴)¹²cos(2x)
. . . . . . . = 2⁴⁸cos(2x)
If we differentiate 2 more times (follow steps A and B)
d⁴⁹y/dx⁴⁹ = 2*2⁴⁸sin(2x) = 2⁴⁹sin(2x)
d⁵⁰y/dx⁵⁰ = 2 * (2⁴⁹cos(2x)) = 2⁵⁰cos(2x)
Of course you don’t have to write out all that – just be able to see and apply the pattern
There is no general method for finding high order derivative – you have to find a pattern

y=cos(2x)
=>
y ‘=2sin(2x)=
2cos[pi/2)+2x]
=>
y”=(2^2)cos(2x)=
(2^2)cos[2(pi/2)+2x]
=>
y”‘=2^3sin(2x)=
(2^3)cos[(3pi/2)+2x]
————–
=>
y^(n)=[2^(n)]cos[(npi/2)+2x]
=>
y^(50)=[2^(50)]cos[25pi+2x]
=>
y^(50)= [2^(50)]cos(2x)

That D^n formula for cos ax is nice, and there’s a similar rule for sin ax, but there really isn’t anything like that for the other trig functions. It’s based on the property that (cos x)’ is just a shifted version of cos x:
(cos x)’ = sin x = cos (x – pi/2)
(cos ax)’ = sin ax * a = a cos (ax – pi/2)
[cos (ax+b)]’ = sin (ax+b) * a = a cos(ax + b – pi/2)
let b = n*pi/2 and
[a^n cos (ax – n*pi/2)]’ = a^n * a cos(ax – n*pi/2 – pi/2)
= a^(n+1) cos [ax – (n+1)*pi/2]
That’s the core of an inductive proof of the formula. When n=0 that amounts to:
(cos ax)’ = a cos (ax – pi/2)
…and that’s the “base case” for the proof by induction.
The other trig functions don’t have such a simple relation between f(x) and f'(x), so taking nth derivatives is very much messier.

i) y = cos(2x) [Given]
ii) Differentiating, y₁ = 2*sin(2x) = 2*cos(π/2 + 2x), since cos(π/2 + u) =sin(u), by complementary relation
Differentiating the above again, y₂ = 2*2*sin(π/2 + 2x) = 4*cos{(π/2) + (π/2 + 2x)} = 4*cos{2*(π/2) + 2x}
= 2²*cos{2*(π/2) + 2x}
In similar lines of the above, differentiating again,
y₃ = 2³*cos{3*(π/2) + 2x}
So of the above, nth derivative = (2^n)*cos{n*(π/2) + 2x}
iii) Hence for 50th derivative, substitute, n = 50 in the above,
50th derivative = (2^50)*cos{50*(π/2) + 2x} = (2^50)*cos{(25π) + 2x} = (2^50)*cos(2x)