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bii17 Group Title

find dy/dx.. y= 3/8 (x) + 3/8 (sinxcosx) + 1/4 (cos^3 x sinx)

  • 2 years ago
  • 2 years ago

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  1. bii17 Group Title
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    \[\frac{ 3 }{ 8 }x + \frac{ 3 }{ 8 }sinxcosx + \frac{ 1 }{ 4 }\cos^3x sinx\]

    • 2 years ago
  2. ali110 Group Title
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    • 2 years ago
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  3. bii17 Group Title
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    the final answer should be cos^4 x

    • 2 years ago
  4. hartnn Group Title
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    it is indeed cos^4 x still interested in finding how ?

    • 2 years ago
  5. bii17 Group Title
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    yep.. can u help me @hartnn

    • 2 years ago
  6. hartnn Group Title
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    sure, u know the product rule in differentiation ?

    • 2 years ago
  7. bii17 Group Title
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    yes..

    • 2 years ago
  8. hartnn Group Title
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    then tell me what u got the derivative of \(cos^3x sin x=?\)

    • 2 years ago
  9. bii17 Group Title
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    \[(\cos^3x) (\cos x) + (\sin x)(3)(\cos^2 x)(-\sin x)\]

    • 2 years ago
  10. hartnn Group Title
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    that is correct :) now u know the formula of sin 2x ?

    • 2 years ago
  11. bii17 Group Title
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    \[\sin 2x= 2sinxcosx\]

    • 2 years ago
  12. bii17 Group Title
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    am i right?

    • 2 years ago
  13. hartnn Group Title
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    yes! u see how we can use that to differentiate middle term easily... 3/8 sin x cos x = 3/16 sin2x after diff. 6/16 cos2x = 3/8 cos 2x ok with this ?

    • 2 years ago
  14. bii17 Group Title
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    where did u get 6/16 cos 2x??

    • 2 years ago
  15. hartnn Group Title
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    \(\huge \frac{d}{dx}sin2x=2cos2x\)

    • 2 years ago
  16. bii17 Group Title
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    ok.. and then.. what's next??

    • 2 years ago
  17. hartnn Group Title
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    the first term diff. is just 3/8. now the 'differentiation' part is over.Lets put it all together and see what we can simplify: \(\large (3/8)+3/8cos2x-(3/4)cos^2xsin^2x+(1/4)cos^4x\) now do u know the formula for cos2x ??

    • 2 years ago
  18. bii17 Group Title
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    there are 3 formulas for cos 2x

    • 2 years ago
  19. hartnn Group Title
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    right, lets get everything in terms of cos....so which formula has cos in it ?

    • 2 years ago
  20. bii17 Group Title
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    cos 2x= cos^2 x - sin^2 x cos 2x= 2cos^2 x - 1 these 2 formulas have cos

    • 2 years ago
  21. hartnn Group Title
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    lets use 2nd one, where we directly get \(1+cos2x=2cos^2x\) and let use the fact that \(sin^2x=1−cos^2x\) so now we have : \((3/4)cos^2x−(3/4)cos^2x(1−cos^2x)+(1/4)cos^4x\) can u simplify this and tell me what gets cancelled

    • 2 years ago
  22. bii17 Group Title
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    hmmm...can i cancell both (3/4)cos^2x ???

    • 2 years ago
  23. bii17 Group Title
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    hmm ok i'll get it ... cos^4 x will be left.

    • 2 years ago
  24. hartnn Group Title
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    but did u get how i got that step ? ask if there is any doubt in any step/term.

    • 2 years ago
  25. bii17 Group Title
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    yes.. I get it.. but about this.. sin^2 x = 1- cos^2 x where din u get it?/

    • 2 years ago
  26. hartnn Group Title
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    basic trignometric identity \(\huge \color{red}{sin^2x+cos^2x=1}\)

    • 2 years ago
  27. bii17 Group Title
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    ow... Thank You For your help :))

    • 2 years ago
  28. hartnn Group Title
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    welcome :)

    • 2 years ago
  29. bii17 Group Title
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    ^^

    • 2 years ago
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