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basic differentiation - is this right?

ok i have this problem here.

y = (sinx)(cosx)(x²+1)

in order to apply product rule here, i need to split it up

y = [(sinx)(cosx)](x²+1)

heres my working to find the derivative

y' = [(sinx)(-sinx)+(cosx)(cosx)](x²+1)

y' = [-sin²x + cos²x](x²+1)

y' = -(1-cos²x)(x²+1)

now i applied the product rule again

y' = -(1-cos²x)(2x)+(x²+1)(2cosx)

y' = -(2x-(-2xcos²x))+(2x²cosx+2cosx)

y' = -(2x+2xcos²x)+2x²cosx+2cosx)

y' = -2x-2xcos²x + 2x²cosx + 2cosx.........which i dont think can be simplified further?

so my answer was that the derivative of (sinx)(cosx)(x²+1) is -2x-2xcos²x + 2x²cosx + 2cosx.........am i correct?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: basic differentiation - is this right?

I think you're confusing yourself a bit.
Splitting it up as you have, you get that:

y' = [(cos x)(sin x)]' (x²+1) + [(cos x)(sin x)]' (x²+1)'

You managed to differentiate (cos x)(sin x) correctly, so the whole thing becomes:

y' = (cos² x - sin² x)(x²+1) + 2x[(cos x)(sin x)]

It would probably have been easier to change (sin x)(cos x) into sin(2x)/2. Then you only have a product of two functions, and so you just use the product rule once without any other complications.

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