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bmelyk

  • 2 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)

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  1. bmelyk
    • 2 years ago
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    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
    • 2 years ago
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    but my answer should look something like: 4cos(x^4)^2*x^4 -4sin(x^4)^2*x^4

  3. satellite73
    • 2 years ago
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    product rule and chain rule

  4. theloser
    • 2 years ago
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    first differentiate sin(x^4) and then x^4.. do the same for cos also

  5. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    similarly if \(g(x)=\cos(x^4)\) then \(g'(x)=-\sin(x^4)\times 4x^3=-4x^3\sin(x^4)\)

  7. theloser
    • 2 years ago
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    @satellite73 said... there will be no x^4 term it is x^3 term

  8. bmelyk
    • 2 years ago
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    i thought that sin and x^4 were two different factors.

  9. bmelyk
    • 2 years ago
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    ?

  10. bmelyk
    • 2 years ago
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    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
    • 2 years ago
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    ok for simplicity.. do one thing.. let some y = x^4

  12. theloser
    • 2 years ago
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    now your equation will be sin(y)cos(y)..

  13. bmelyk
    • 2 years ago
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    ok

  14. bmelyk
    • 2 years ago
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    so i get: siny'cosy+cosy'-siny

  15. theloser
    • 2 years ago
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    you know how to differentiate this right.. that will (cos(y)^2-sin(y)^2)dy

  16. bmelyk
    • 2 years ago
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    yes

  17. bmelyk
    • 2 years ago
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    i do the product as: fg'+gf'

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

  20. theloser
    • 2 years ago
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    now our y = x^4.. so y'= (4x^3) x'

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

  22. bmelyk
    • 2 years ago
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    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
    • 2 years ago
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    what you have seems to look like that answer though is it?

  24. theloser
    • 2 years ago
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    ok.. how did you get 4terms?

  25. theloser
    • 2 years ago
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    mistake you are would probably be this.. you are combining both product rule and chain rule..

  26. bmelyk
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    yes i have that done.

  29. bmelyk
    • 2 years ago
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    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
    • 2 years ago
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    yes.. that is what you should get..

  31. bmelyk
    • 2 years ago
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    can i write it as: 4x^3*cos(x^4)^2-sin(x^4)^2

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