find derivative of sin(x)sin(2x)sin(3x)?

Question

our professory wants us to use ((x+h) -f(x))/h but anything will help

Answer 1

Rewrite sin(x)sin(2x)sin(3x) using the sinAsinB formula on the first two terms.
This gives:
sin(x)sin(2x)sin(3x)
= 1/2(cosx - cos3x) sin 3x
=1/2 cosxsin3x -1/2 cos3xsin3x
Now use the sinAcosB expansion to give
=1/4(sin2x + sin4x - sin6x)

You can then use the standard limit definition for the derivative of sinx on each of these three terms.

ie f'(x) = lim h->0 (f(x+h) -f(x))/h
= lim h->0 1/4(sin2(x +h) -sin2x)/h)
=1/4 lim h->0 (sin2xcos2h + cos2xsin2h -sin2x)/h
etc actually I've forgotten how it goes from there but a bit of fiddling should get it out!!

No, that's wrong use the Sums to products on the top line - ie sinA - sinB = 2cos(half A+B)sin(half A-B)

The guy above is correct if you just want the answer. Derivative of fgh = f'gh +fg'h + fgh'

Answer 2

Differentiate Sin2x

Answer 3

Since your professor asked you to use the limit definition of derivatives, I'd suggest using the trig identity
sin(a+b)=sin(a)cos(b) + sin(b)cos(a)
and play with it. Also, don't forget to use the special limit identity
lim(h->0) sin(h)/h = 1
Since this is your homework, I cannot give you more information.

It'd be long and tedious. Good luck.

Answer 4

well the product rule for derivative is g'(x)h(x) + h'(x)g(x) so i guess you could do sin(2x)sin(2x) first and then use the product and multiply that by sin(3x) OR you could go find those pesky trig identities somewhere, multiply out the whole thing to get one function, and use (f(x+h)-f(x))/h. just a thought, though. not entirely sure how it would work out.

Answer 5

The derivative of f(x)g(x)h(x) is:

f'(x)g(x)h(x) + f(x)g'(x)h(x) + f'(x)g(x)h'(x)

In this case the derivative of (sin x)(sin 2x)(sin 3x) is:

(cos x)(sin 2x)(sin 3x) + 2(sin x)(cos 2x)(sin 3x)
+ 3(sin x)(sin 2x)(cos 3x)

Answer 6

first keep sinx outside and differentiate sin2x.sin3x by chain rule.then keep sin2x outside &differentiate.then do the same with sin3x