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anonymous

  • one year ago

Can someone please help me find the derivative of this problem? I will fan and medal.

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  1. Jhannybean
    • one year ago
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    Whats the function?

  2. anonymous
    • one year ago
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    I'm typing it

  3. anonymous
    • one year ago
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    \[\frac{ 2 }{ 3 }\sin^\frac{ 3 }{ 2 }x-\frac{ 2 }{ 7 }\sin^\frac{ 7 }{ 2 }x\]

  4. Jhannybean
    • one year ago
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    \[\frac{2}{3}\left(\sin^{2/3}x\right)' \iff \frac{2}{3}\left((\sin(x))^{2/3}\right)'\]

  5. wmj259
    • one year ago
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    If you look carefully you see that the two functions are connected with the subtract sign. So you can break that function into 2 separate derivatives.

  6. wmj259
    • one year ago
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    \[\frac{ d }{ dx } (\frac{ 2 }{ 3 }\sin ^{2/3}(x)) - \frac{ d }{ dx }((\frac{ 2 }{ 7 }\sin ^{7/2}x))\]

  7. Jhannybean
    • one year ago
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    Use the power rule, then the chain rule.

  8. wmj259
    • one year ago
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    |dw:1444350991307:dw|

  9. Jhannybean
    • one year ago
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    lmao -_-

  10. anonymous
    • one year ago
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    Wait. How did you guys turn the exponent 3/2 into 2/3?

  11. Jhannybean
    • one year ago
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    Typo.

  12. wmj259
    • one year ago
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    yes that was a typo on my part, I apologize.

  13. anonymous
    • one year ago
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    Y'all are fine. I was just making sure.

  14. Jhannybean
    • one year ago
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    \[\large \frac{d}{dx}\left[\frac{2}{3}(\sin(x))^{3/2}\right] = \frac{2}{3} \cdot \frac{3}{2} \cdot (\sin(x))^{(3/2) - 1} \cdot \frac{d}{dx} (\sin(x))\]

  15. wmj259
    • one year ago
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    What I would do in this situation would be to first split the problem into 2 different differentiation's because of that subtract sign. You can pull out the (2/3) and the (2/7) because they are constants And then I would do the Power Rule on the (3/2) and the (7/2). Separately. And then I would do the chain rule on the 2 sin(x) functions.

  16. wmj259
    • one year ago
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    |dw:1444351292444:dw|

  17. Jhannybean
    • one year ago
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    \[\large \frac{d}{dx}\left[\frac{2}{7}(\sin(x))^{7/2}\right] = \frac{2}{7} \cdot \frac{7}{2} \cdot (\sin(x))^{(7/2)-1} \cdot \frac{d}{dx}(\sin(x))\]

  18. wmj259
    • one year ago
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    |dw:1444351320870:dw|

  19. wmj259
    • one year ago
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    |dw:1444351344236:dw|

  20. anonymous
    • one year ago
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    \[\sin^\frac{ 1 }{ 2 }x*cosx-\sin^\frac{ 5 }{ 2 }x*cosx\]

  21. anonymous
    • one year ago
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    That's what I got

  22. Jhannybean
    • one year ago
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    For which part?

  23. anonymous
    • one year ago
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    The entire problem

  24. Jhannybean
    • one year ago
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    Alright let's check

  25. anonymous
    • one year ago
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    I just did the power rule then chain rule as recommended.

  26. wmj259
    • one year ago
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    |dw:1444351512241:dw|

  27. wmj259
    • one year ago
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    Is the finished derivative for the left half.

  28. wmj259
    • one year ago
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    the simplified left side is |dw:1444351605530:dw|

  29. wmj259
    • one year ago
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    Now the right side would give us: |dw:1444351641222:dw|

  30. wmj259
    • one year ago
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    The simplified right side would be |dw:1444351722466:dw|

  31. wmj259
    • one year ago
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    So if you combine the left and the right side you get: |dw:1444351765027:dw|

  32. wmj259
    • one year ago
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    @Catseyeglint911, Which is what you got.

  33. anonymous
    • one year ago
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    @wmj259 Can you simplify the right side any further?

  34. wmj259
    • one year ago
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    you can say this: |dw:1444351917329:dw|

  35. wmj259
    • one year ago
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    |dw:1444351988066:dw|

  36. Jhannybean
    • one year ago
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    \[\begin{align}\large \frac{d}{dx}\left[\frac{2}{3}(\sin(x))^{3/2}\right] &= \frac{2}{3} \cdot \frac{3}{2} \cdot (\sin(x))^{(3/2) - 1} \cdot \frac{d}{dx} (\sin(x)) \\&= \sin^{1/2}(x)\cos(x) \ \end{align}\]\[\begin{align} \large \frac{d}{dx}\left[\frac{2}{7}(\sin(x))^{7/2}\right] &= \frac{2}{7} \cdot \frac{7}{2} \cdot (\sin(x))^{(7/2)-1} \cdot \frac{d}{dx}(\sin(x)) \\&=\sin^{5/2}(x)\cos(x)\end{align}\]\[\begin{align}\frac{d}{dx}\left[ \frac{ 2 }{ 3 }\sin^\frac{ 3 }{ 2 }x-\frac{ 2 }{ 7 }\sin^\frac{ 7 }{ 2 }x\right] &= \sin^{1/2}(x)\cos(x) -\sin^{5/2}(x)\cos(x) \\&= \sin^{1/2}(x)\cos(x) \left[ 1-\sin^{5}(x) \right] \end{align} \]

  37. Jhannybean
    • one year ago
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    \[\large \color{red}{\boxed{\sin^{1/2}(x)\cos(x)\left[1-\sin^{5}(x)\right]}}\]

  38. Jhannybean
    • one year ago
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    @wmj259 you did something funky here|dw:1444352122644:dw|

  39. wmj259
    • one year ago
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    Yes, The right most part should have been the derivative of sin(x) not just (x).

  40. wmj259
    • one year ago
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    It goes much longer then that because their is basically a Power rule then 2 chain rules. but the 2nd chain rule is always ignore because its just the derivative of x's in some cases. But with practice comes great speed.

  41. anonymous
    • one year ago
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    Thank you both! I wish I could give out multiple medals. I'll just become a fan of both of you. :)

  42. Jhannybean
    • one year ago
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    Woohoo!

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