Question

Find the nth derivative of sin2xsin4x

Answer 1

Here, f(x) = sin 2x sin 4x 
sin(a)sin(b) = (1/2) ( cos(a-b) - cos(a+b)) 
sin 2x sin 4x = (1/2) ( cos(-2x) - cos(6x) ) = (1/2) cos(2x) - (1/2) cos(6x) 

f'(x) = (1/2) (-2 sin 2x) - (1/2) (-6 sin 6x) 
f'(x) = -sin 2x + 3 sin 6x 
f''(x) = - 2 cos 2x + 18 cos 6x 
f'''(x) = 4 sin 2x - 108 sin 6x 
f^4(x) = 8 cos 2x - 648 cos 6x 
f^5(x) = -16 sin 2x + 3,888 sin 6x 
f^6(x) = -32 cos 2x + 23,328 cos 6x 
f^7(x) = 64 sin 2x - 139,968 sin 6x 

Now, the odd derivatives have the sum of two sine terms 
and, the even derivatives have the sum of two cosine terms 

So, nth derivative: 

If n is even: 
(-1)^(n-1) [ 2^(n-1) cos 2x - 3*6^(n-1) cos 6x ] 

If n is odd: 
f^n(x) = [-2^(n-1) sin 2x + 3*6^(n-1) sin 6x] 
or 
f^n(x) = [2^(n-1) sin 2x - 3*6^(n-1) sin 6x]

Answer 2

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