# Finding 50th derivative of y = cos2x?? HELP ASAP?

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!

• 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 A-D, 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)