Differentiation

[roars]

I know this is not code related but i use this forum for vb and found it very good so hopefully i can get help with this problem!

I am doing a repeat module in uni which i failed over a year ago and i am having problems getting back into it! Heres my problem:

Show that the derivative of f(x) = (sin x)^3 + (cos x)^2 *(sin x) + 3x^2 = cosx +6x (where the 2 & 3 are sin squared and cos cubed respectively)

I can get this far but not sure what to do next!

(sin x)^3 = 3((sin x)^2)*(cos x)

(cos x)^2 = -2(cos x)(sin x)

sin x = cos x

therefore i get

3((sin x)^2)*(cos x) + (cos x)^3 + (sin x) -2(cos x)(sin x) + 6x

where do i go from here?

Thank You
Ruairi

[thinktank2]

3((sin x)^2)*(cos x) + (cos x)^3 + (sin x) -2(cos x)(sin x) + 6x

= 3[(Sin x)^2 * (Cos x)] + (Cos x)^3 - 2[(Sin x)^2 * (Cos x)] + 6x

= (3-2)[(Sin x)^2 * (Cos x)] + (Cos x)^3 + 6x

= [(Sin x)^2 * (Cos x)] + (Cos x)^3 + 6x

= [(1 - (Cosx)^2) * (Cos x)] + (Cos x)^3 + 6x

= (Cosx) - (Cosx)^3 + (Cos x)^3 + 6x

= (Cosx) + 6x

[Ambivalentiowa]

Originally posted by roars

sin x = cos x

Where did you get that from?

[roars]

Sorry I maybe didnt explain myself well enough there, for those 3 expressions on the left I have put their differential on the right.

Should probably be:

dy/dx sin x = cos x

Hope this makes it clearer.

Thanks for the reply thinktank helped a lot!