## bmelyk Group Title Use the appropriate rule or combination of rules to find the derivative of the function defined below. y = sin (x^4) cos (x^4) one year ago one year ago

1. bmelyk Group Title

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 Group Title

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

3. satellite73 Group Title

product rule and chain rule

4. theloser Group Title

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

5. satellite73 Group Title

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 Group Title

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

7. theloser Group Title

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

8. bmelyk Group Title

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

9. bmelyk Group Title

?

10. bmelyk Group Title

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 Group Title

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

12. theloser Group Title

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

13. bmelyk Group Title

ok

14. bmelyk Group Title

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

15. theloser Group Title

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

16. bmelyk Group Title

yes

17. bmelyk Group Title

i do the product as: fg'+gf'

18. theloser Group Title

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 Group Title

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

20. theloser Group Title

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

21. theloser Group Title

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

22. bmelyk Group Title

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 Group Title

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

24. theloser Group Title

ok.. how did you get 4terms?

25. theloser Group Title

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

26. bmelyk Group Title

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 Group Title

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 Group Title

yes i have that done.

29. bmelyk Group Title

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 Group Title

yes.. that is what you should get..

31. bmelyk Group Title

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