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anonymous

  • 5 years ago

if i am looking for the derivative of f(x)=sin2x(3x+7)^5 would i come out with f'(x)=5(sin2x)^4(cos2)(6x+21)

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  1. anonymous
    • 5 years ago
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    What do you guys think?

  2. anonymous
    • 5 years ago
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    wrong, way off

  3. anonymous
    • 5 years ago
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    there are many chain rules here...make sure you apply them

  4. anonymous
    • 5 years ago
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    and i'm assuming that's sin(2x) (3x+7)^5?

  5. anonymous
    • 5 years ago
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    review your product rules...and go slowly =)

  6. anonymous
    • 5 years ago
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    =sin2x du/dx = 2cos(2x) v= (3x+7)^5 dv/dx = 5 (3x+7)^4 * 3 = 15 (3x+7)^4 now product rule

  7. anonymous
    • 5 years ago
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    Find the derivative of the function f(x) = sin 2x(3x + 7)^5

  8. anonymous
    • 5 years ago
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    f'(x) = 2cos(2x)(3x+7)^5 + 15sin(2x) (3x+7)^4

  9. anonymous
    • 5 years ago
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    thats copied from the guide

  10. anonymous
    • 5 years ago
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    what elec wrote is correct...so please do it yourself and compare with what he wrote...there's no point copying the answer...make sure you know how he derived it

  11. anonymous
    • 5 years ago
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    Thanks for the help

  12. anonymous
    • 5 years ago
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    np, let me know if something seems fishy...it's better to do the leg work now....or you'll regret it later =)

  13. anonymous
    • 5 years ago
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    Isn't the derivative of cos = -sin?

  14. anonymous
    • 5 years ago
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    it is...but you are looking for the derivative of sine which is cosine

  15. anonymous
    • 5 years ago
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    so... d/dx sin(x) = cos(x) d/dx cos(x) = -sin(x)

  16. anonymous
    • 5 years ago
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    ok i see that now thanks, can you break down how he got the 15sin it looks to me that it would be 10sin(2x)(3x+7)^4

  17. anonymous
    • 5 years ago
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    I may just be really really confused now

  18. anonymous
    • 5 years ago
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    careful there you shouldn't be doing a chain rule on the sin(2x) ...you need to do it on the (3x+7)...so actually...you get 15sin(2x)(3x+7)^4

  19. anonymous
    • 5 years ago
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    it might help to break up your function... f(x) = sin(2x) * (3x + 7)^5 = u(x) * v(x) f'(x) = u'(x)*v(x) + v'(x)*u(x) try to take it from here ( I let u(x) = sin(2x) and v(x) = (3x+7)^5 )

  20. anonymous
    • 5 years ago
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    f'(x)=2cos2*(3x+7)^5+15x*sin(2x)(3x+7)^4

  21. anonymous
    • 5 years ago
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    ok i caught it that time, thanks for the help

  22. anonymous
    • 5 years ago
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    you are welcome

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