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nickymarden

  • 3 years ago

derive the equation.

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  1. nickymarden
    • 3 years ago
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  2. Rogue
    • 3 years ago
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    Senx? What kind of math is this? :o

  3. nbouscal
    • 3 years ago
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    Is that supposed to be sin x?

  4. nickymarden
    • 3 years ago
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    sorry, it's in portugese :) and yes it is.

  5. Rogue
    • 3 years ago
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    Probably. But what is the question asking for? Derive it from...?

  6. nbouscal
    • 3 years ago
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    Okay, so \(g(x)=(\sin x+\sqrt{x})^3\cdot\sqrt{x^4+2}\) and you're looking for \(g'(x)\)?

  7. lgbasallote
    • 3 years ago
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    no wonder i have seen these sen x for a lot of times i thought they taught secret trigonometry to people -___-

  8. nbouscal
    • 3 years ago
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    So, use the product rule and the chain rule, yeah? Which part are you getting stuck at?

  9. nickymarden
    • 3 years ago
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    i'm sorry for taking long. I was tryig to do it by myself. I just can't do it lol

  10. nbouscal
    • 3 years ago
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    Show us your work so we can see where you're getting lost.

  11. nbouscal
    • 3 years ago
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    First step, product rule, right? So \(f(x)=(\sin x+\sqrt{x})^3, g(x)=\sqrt{x^4+2}\), and we're finding \(f'(x)g(x)+g'(x)f(x)\)

  12. nickymarden
    • 3 years ago
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    yeeah

  13. Rogue
    • 3 years ago
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    \[g(x) = (\sin x + \sqrt x)^3 *\sqrt {x^4 +2}\]This gets messy...\[\frac {dg}{dx} = (\sin x + \sqrt x)^3 \frac {d}{dx} \left[ \sqrt {x^4 +2} \right] + \sqrt {x^4 +2} \frac {d}{dx} \left[ (\sin x + \sqrt x)^3 \right]\]

  14. Rogue
    • 3 years ago
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    As nbouscal said, where do you get stuck?

  15. nickymarden
    • 3 years ago
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    yeah, i get lost in all the algebra.

  16. nickymarden
    • 3 years ago
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    i just erased it all, let me try again lol.

  17. nbouscal
    • 3 years ago
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    Take it one step at a time. First step is to find f'(x), so derive \((\sin x + \sqrt x)^3\) using the chain rule.

  18. nickymarden
    • 3 years ago
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    thank you for the help :))

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