(sin5x)(sin6x)(cos11x)

Question

Find the nth derivative of the following

Answer 1

cos(A-B)=cosAcosB+sinAsinB; cos(A+B)=cosAcosB-sinAsinB;

cos(A-B)-cos(A+B)=2sinAsinB.

Let A=6x and B=5x, then A-B=x and A+B=11x, and cos(x)-cos(11x)=2sin(6x)sin(5x).

sin(6x)sin(5x)=½(cos(x)-cos(11x));

sin(6x)sin(5x)sin(11x)=½(cos(x)-cos(11x))sin(11x)=½cos(x)sin(11x)-½sin(11x)cos(11x),

sin(6x)sin(5x)sin(11x)=½cos(x)sin(11x)-¼sin(22x).

sin(A+B)=sinAcosB+cosAsinB; sin(A-B)=sinAcosB-cosAsinB,

sin(A+B)+sin(A-B)=2sinAcosB.

Let A=11x and B=x, then sin(12x)+sin(10x)=2sin(11x)cos(x), therefore:

sin(6x)sin(5x)sin(11x)=½cos(x)sin(11x)-¼sin(22x)=¼(sin(12x)+sin(10x)-sin(22x).

Let y₁=¼sin(12x), y₂=¼sin(10x), y₃=-¼sin(22x), so y= y₁+y₂+y₃.

The derivative of the sum y is the sum of the derivatives of y₁, y₂ and y₃.

y₁'=3cos(12x), y₂'=(5/2)cos(10x), y₃'=-(11/2)cos(22x),

y₁"=-36sin(12x)=-144y₁,  y₂"=-25sin(10x)=-100y₂, y₃"=121sin(22x)=-484y₃.

This gives us a pattern for higher derivatives.

Let integer n≥0 denote the degree of the derivative: y⁽ⁿ⁾, where n=0 implies y, y⁽¹⁾=y'=dy/dx, etc.

Now consider 4 different values of n: 4k, 4k+1, 4k+2, 4k+3 where integer k≥0.

Consider derivatives of y₁ as an example:

y₁⁽⁰⁾=¼sin(12x), y₁⁽¹⁾=3cos(12x),

y₁⁽²⁾=-12²y₁⁽⁰⁾, y₁⁽³⁾=-12²y₁⁽¹⁾

y₁⁽⁴⁾=-12²y₁⁽²⁾=12⁴y₁⁽⁰⁾=(12⁴/4)sin(12x),

y₁^(4k)=12^4ky₁⁽⁰⁾=(12^4k/4)sin(12x)=((3×4)^4k/4)sin(12x)=(3^4k4^4k-1)sin(12x);

y₁^(4k+1)=(3^4k4^4k-1)(12cos(12x)=(3^4k+14^4k)cos(12x);

y₁^(4k+2)=-(3^4k+24^4k+1)sin(12x);

y₁^(4k+3)=-(3^4k+34^4k+2)cos(12x).

The same rules apply mutatis mutandis to the other y's:

y₂^(4k)=(5^4k2^4k-2)sin(10x); y₂^(4k+1)=(5^4k+12^4k-1)cos(10x);

y₂^(4k+2)=-(5^4k+22^4k)sin(10x); y₂^(4k+3)=-(5^4k+32^4k+1)cos(10x).

y₃^(4k)=-(11^4k2^4k-2)sin(22x); y₃^(4k+1)=-(11^4k+12^4k-1)cos(22x);

y₃^(4k+2)=(11^4k+22^4k)sin(22x); y₃^(4k+3)=(11^4k+32^4k+1)cos(22x).

So we can write:

y^(4k)=(3^4k4^4k-1)sin(12x)+(5^4k2^4k-2)sin(10x)-(11^4k2^4k-2)sin(22x);

y^(4k+1)=(3^4k+14^4k)cos(12x)+(5^4k+12^4k-1)cos(10x)-(11^4k+12^4k-1)cos(22x);

y^(4k+2)=-(3^4k+24^4k+1)sin(12x)-(5^4k+22^4k)sin(10x)+(11^4k+22^4k)sin(22x);

y^(4k+3)=-(3^4k+34^4k+2)cos(12x)-(5^4k+32^4k+1)cos(10x)+(11^4k+32^4k+1)cos(22x).