Question

Find the nth
order derivations
y = sinmx + cosmx​

Answer 1

Step-by-step explanation:

If you take the first few derviatives you will see a pattern.

y ' = m (cos(mx) - sin(mx)

y '' = m^2 (-sin(mx) - cos(mx))

y ''' = m^3 (-cos(mx) + sin(mx))

y '''' = m^4 (sin(mx) + cos(mx))

After this the pattern of sin(mx) and cos(mx) repeats, and you just add another

power of m. So you can see where the m^n comes from in the assertion.

Now you just need to figure out what is going on with the rest of it.

Consider what happens when n is 1. You;d have

[1 + (-1)^1sin(2mx)]^1/2

replace the 1 with sin^2(mx) + cos^2(mx) and

replace sin(2mx) with 2sin(mx)cos(mx) and you get

[sin^2*(mx) + cos^2(mx) -2sin(mx)cos(mx)]^1/2

reorder this to get

[sin^2(mx) - 2sin(mx)cos(mx) + cos^2(mx)]^1/2

recognize that this is a (x-y)^2 = x^2 -2xy + y^2 situation

[ (sin(mx) - cos(mx))^2]^1/2 This can be simplied to either

sin(mx) - cos(mx) or - (sin(mx) - cos(mx))

If we take the 2nd version of this this matches our result for y '

Do this same this for y'' and the you see that the -1^n term flips the sign for the

sin(2mx) term. Eventually you end up with

[ sin^2(mx) + 2sin(mx) cos(mx) + cos^2(mx)]^1/2

which can then be simplified to either

sin(mx) + cos(mx) or - (sin(mx) +cos(mx))

The 2nd verson of this gets you the correct result for y ''

If you look at the y '''' case you end up back at a choice of

sin(mx) - cos(mx) or - (sin(mx) - cos(mx))

But this time you want the first version

and when you do y ' ' ' ' you get back to a choice of

sin(mx) + cos(mx) or - (sin(mx) +cos(mx))

and this time you need the first version.

And then the pattern repeats