You are not logged in.
Pages: 1
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?
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.
Why did the vector cross the road?
It wanted to be normal.
Offline
Pages: 1