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anonymous

  • 5 years ago

find the derivative: (3x+1)^3/(1-3x)^4 I get the answer (3x+1)^2+(9x+21)/(1-3x)^5 is this correct?

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  1. anonymous
    • 5 years ago
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    the + between the 2 equations in the answer should not be there should be (3x+1)^2 (9x+21)/(1-3x)^5

  2. amistre64
    • 5 years ago
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    [ 9 (1-3x)^4 (3x+1)^2 ] - [ -12 (3x+1)^3 (1-3x)^3 ] ---------------------------------------------- (1-3x)^8

  3. anonymous
    • 5 years ago
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    i have that much

  4. amistre64
    • 5 years ago
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    just working it out in me head :)

  5. amistre64
    • 5 years ago
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    9bt' - (-12b't) 9bt' + 12b't ------------ = ------------ b^2 b^2 the (+) is good

  6. anonymous
    • 5 years ago
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    you lost me on that one

  7. amistre64
    • 5 years ago
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    we get 2 terms up top that are made up of multiplication. the first term has a (+)9 as a constant, and the 2nd term has a (-)12 as a constant. 9 --12 = 9+12

  8. anonymous
    • 5 years ago
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    ok I got that. I guess the way it was written confussed me

  9. amistre64
    • 5 years ago
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    probably, I was just trying to clean it up for my eyes :)

  10. amistre64
    • 5 years ago
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    for simplicity: r = (1-3x) ; s = (3x+1) 3 r^3 s^2 (3r + 4s) ------------------ r^8

  11. anonymous
    • 5 years ago
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    Let f(x) and g(x) be the Numerator and the Denominator of the given fraction respectively. The derivative of the fraction in terms of the above functions is:\[\frac{f'[x]}{g[x]}-\frac{f[x] g'[x]}{g[x]^2} \] Plug in the function values and their associated derivatives and you should get: \[\frac{9 (1+3 x)^2}{(1-3 x)^4}+\frac{12 (1+3 x)^3}{(1-3 x)^5} \]

  12. amistre64
    • 5 years ago
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    rob: I dont follow that.....

  13. amistre64
    • 5 years ago
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    might be right, but im lost on it :)

  14. anonymous
    • 5 years ago
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    To start from the beginning. mom wants to know if the derivative of \[{(3x+1)^3 \over {(1-3x)^4}} \] is equal to: \[(3 x+1)^2+\frac{(9 x+21)}{(1-3 x)^5} \] The answer is no. The derivative is: \[\frac{9 (1+3 x)^2}{(1-3 x)^4}+\frac{12 (1+3 x)^3}{(1-3 x)^5} \] Let \[\frac{(3 x+1)^3}{(1-3 x)^4}=\frac{f(x)}{g(x)} \] The derivative of f(x)/g(x) is \[\frac{f'(x)}{g(x)}-\frac{f(x) g'(x)}{g(x)^2} \] \[f(x) = (1+3x)^3, f'(x) = 9(1+3x)^2 \] \[g(x) = (1-3x)^4, g'(x) = -12(1-3x)^3\] Plug in the values for f(x), f'(x), g(x) and g'(x) into the derivative of the f(x)/g(x) and one should end up with the equivalent of the derivative. \[\frac{9 (1+3 x)^2}{(1-3 x)^4}-\frac{(1+3 x)^3 \left(-12 (1-3 x)^3\right)}{\left((1-3 x)^4\right)^2} \] or \[\frac{9 (1+3 x)^2}{(1-3 x)^4}+\frac{12 (1+3 x)^3}{(1-3 x)^5} \]

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