## bmelyk 3 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. bmelyk

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. bmelyk

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

3. satellite73

product rule and chain rule

4. theloser

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

5. satellite73

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. satellite73

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

7. theloser

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

8. bmelyk

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

9. bmelyk

?

10. bmelyk

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. theloser

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

12. theloser

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

13. bmelyk

ok

14. bmelyk

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

15. theloser

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

16. bmelyk

yes

17. bmelyk

i do the product as: fg'+gf'

18. theloser

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

19. bmelyk

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

20. theloser

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

21. theloser

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

22. bmelyk

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. bmelyk

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

24. theloser

ok.. how did you get 4terms?

25. theloser

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

26. bmelyk

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. theloser

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. bmelyk

yes i have that done.

29. bmelyk

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. theloser

yes.. that is what you should get..

31. bmelyk

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