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

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 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)