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marmots

  • 4 years ago

if f(x)= (3sqrtX)(X^3 -2sqrtX +6) find f'(x)

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  1. amistre64
    • 4 years ago
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    do we assume X = x ?

  2. marmots
    • 4 years ago
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    yes

  3. amistre64
    • 4 years ago
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    product rule it then and youll be good

  4. jrodriguez2315
    • 4 years ago
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    Take the derivative of f'(x). Use product rule along with chain rule. f'(x)=f'(x)g(x) + f(x)g'(x).

  5. marmots
    • 4 years ago
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    I've tried that several times. I just need an answer.

  6. amistre64
    • 4 years ago
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    (3sqrtX)'(X^3 -2sqrtX +6)+(3sqrtX)(X^3 -2sqrtX +6)' \(\cfrac{3}{2\sqrt{x}}\) (X^3 -2sqrtX +6)+ \(3\sqrt{x}\ (3x^2 -\cfrac{1}{\sqrt{x}})\)

  7. amistre64
    • 4 years ago
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    \[\frac{3x^3}{2\sqrt{x}}-3+\frac{9}{\sqrt{x}}+9x^2\sqrt{x}-3\]

  8. jrodriguez2315
    • 4 years ago
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    marmots make sure to regard the sqrt(x) as (x)^1/2. It makes it easier to integrate. Just a tip.

  9. jrodriguez2315
    • 4 years ago
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    I mean to take the derivative!!

  10. marmots
    • 4 years ago
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    Thanks guys, I appreciate it.

  11. amistre64
    • 4 years ago
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    yep, might need to simplify some more, but im pretty sure thats the brunt of it

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