## anonymous 4 years ago Use the appropriate rule or combination of rules to find the derivative of the function defined below. y = sin (x^4) cos (x^4)

1. anonymous

i used the product rule for both sin and cos to get (sin*4x^3+x^4*cos)(cos*4x^3*x^4-sinx)

2. anonymous

but my answer should look something like: 4cos(x^4)^2*x^4 -4sin(x^4)^2*x^4

3. anonymous

product rule and chain rule

4. anonymous

first differentiate sin(x^4) and then x^4.. do the same for cos also

5. anonymous

the answer you wrote second is not right either you need $$(fg)'=f'g+g'f$$ with $f(x)=\sin(x^4), f'(x)=\cos(x^4)\times 4x^3=4x^3\cos(x^4)$

6. anonymous

similarly if $$g(x)=\cos(x^4)$$ then $$g'(x)=-\sin(x^4)\times 4x^3=-4x^3\sin(x^4)$$

7. anonymous

@satellite73 said... there will be no x^4 term it is x^3 term

8. anonymous

i thought that sin and x^4 were two different factors.

9. anonymous

?

10. anonymous

should i end up with something that looks like: $4x^3*\cos(x^4)*\cos(x^4)+(-4sinx^3*\sin(x^4))*\sin(x^4)$

11. anonymous

ok for simplicity.. do one thing.. let some y = x^4

12. anonymous

now your equation will be sin(y)cos(y)..

13. anonymous

ok

14. anonymous

so i get: siny'cosy+cosy'-siny

15. anonymous

you know how to differentiate this right.. that will (cos(y)^2-sin(y)^2)dy

16. anonymous

yes

17. anonymous

i do the product as: fg'+gf'

18. anonymous

yes perfect.. d(uv) =vd(u)+ud(v) right.. v =cos(y) u=sin(y).. from this you will get the above thing

19. anonymous

i don't see how you're getting it to the power of 2 when it should be 1-1. shouldnt it?

20. anonymous

now our y = x^4.. so y'= (4x^3) x'

21. anonymous

put y and y' in above equation.. what do you get.. 4x^3(cos(x^4)^2-sin(x^4)^2)

22. anonymous

i didn't get everything to the power of 2, i just had(sin(4x^3)+cos(x^4))+((cos(4x^3)-sin(x^4)

23. anonymous

what you have seems to look like that answer though is it?

24. anonymous

ok.. how did you get 4terms?

25. anonymous

mistake you are would probably be this.. you are combining both product rule and chain rule..

26. anonymous

i think that is what i'm doing, i thought that is what i neeeded to do. so my answer should be: 4x^3(cos(x^4)^2-sin(x^4)^2)

27. anonymous

see when you differentiate sin(x^4).. first you sin derivative (cos(x^4)*(derivative of x^4))this whole thing is multiplied by cos(x^4).. do the same for cos(x^4) also

28. anonymous

yes i have that done.

29. anonymous

is 4x^3(cos(x^4)^2-sin(x^4)^2) what i should have? i find it easier to have a correct answer so i can look through my workings and see what i've done wrong.

30. anonymous

yes.. that is what you should get..

31. anonymous

can i write it as: 4x^3*cos(x^4)^2-sin(x^4)^2