1. ## Differential problem!

ok, i was doing my hw and i just do not have a clue how to solve this.
the problem is:
(2x^3-x)cos(1-x^2)

and this is what i got or attempted so far. i'm not even sure if this is the right way. can somebody tell me where i went wrong?

[d/dx(2x^3-x)cos)1-x^2) x (2x^3-x)[d/dx(cos(1-x^2))]
=(6x^2-1)cos(1-x^2) x 2x(2x^3-x)sin(1-x^2)

thanks much!

2. Originally Posted by katieeej
ok, i was doing my hw and i just do not have a clue how to solve this.
the problem is:
(2x^3-x)cos(1-x^2)

and this is what i got or attempted so far. i'm not even sure if this is the right way. can somebody tell me where i went wrong?

[d/dx(2x^3-x)cos)1-x^2) x (2x^3-x)[d/dx(cos(1-x^2))]
=(6x^2-1)cos(1-x^2) x 2x(2x^3-x)sin(1-x^2)

thanks much!
Replace the times-sign by -

3. Originally Posted by katieeej
ok, i was doing my hw and i just do not have a clue how to solve this.
the problem is:
(2x^3-x)cos(1-x^2)

and this is what i got or attempted so far. i'm not even sure if this is the right way. can somebody tell me where i went wrong?

[d/dx(2x^3-x)cos)1-x^2) x (2x^3-x)[d/dx(cos(1-x^2))]
=(6x^2-1)cos(1-x^2) x 2x(2x^3-x)sin(1-x^2)

thanks much!
.
No! The product rule: The derivative of fg is NOT f'g' nor is it f'- g'. It is (fg)'= f'g+ fg'.