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anonymous

  • 5 years ago

Use logarithmic differentiation...

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  1. anonymous
    • 5 years ago
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    \[y = \frac{x^{2}\sqrt{3x-2}}{(x+1)^{2}} \] and x > 2/3

  2. myininaya
    • 5 years ago
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    Take ln() of both sides.

  3. myininaya
    • 5 years ago
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    \[\ln(y)=\ln(\frac{x^2 \sqrt{3x-2}}{(x+1)^2})\]

  4. myininaya
    • 5 years ago
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    Use properties of log to expand

  5. anonymous
    • 5 years ago
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    Currently I'm doing this: \[lny = 2lnx+\frac{1}{2}\ln(3x-2)-2\ln(x+1)\]

  6. myininaya
    • 5 years ago
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    ok great

  7. myininaya
    • 5 years ago
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    \[(\ln(f(x)))'=\frac{f'(x)}{f(x)}\]

  8. anonymous
    • 5 years ago
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    and then I differentiate that and move the y over to get: \[\frac{dy}{dx} = y\left[\frac{3x^{2}+15x-8}{2x(3x-2)(x+1)} \right]\] However, in all of my solutions in the manual, they seem to substitute y with y's reciprocal and I don't understand why.

  9. anonymous
    • 5 years ago
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    they give the answer to be:\[\frac{(3x^{3}+15x^{2}-8x)}{2(x+1)^{3}\sqrt{3x-2}}\]

  10. anonymous
    • 5 years ago
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    At least I'm assuming they're using the reciprocal because the sqrt always ends up on the opposite side of the fraction than where it was in the original y (numerator->denominator or vice versa)

  11. myininaya
    • 5 years ago
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    \[y'=y(2 \frac{1}{x}+\frac{1}{2}\frac{3}{3x-2}-2\frac{1}{x+1}) \] \[y' = \frac{x^{2}\sqrt{3x-2}}{(x+1)^{2}}(\frac{2(2)(3x-2)(x+1)}{2x(3x-2)(x+1)}+\frac{3x(x+1)}{2x(3x-2)(x+1)}-\frac{2x(2)(3x-2)}{2x(3x-2)(x+1)})\]

  12. myininaya
    • 5 years ago
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    \[y'=\frac{x^2 \sqrt{3x-2}}{(x+1)^2} \cdot \frac{4(3x-2)(x+1)+3x(x+1)-4x(3x-2)}{2x(3x-2)(x+1)}\]

  13. myininaya
    • 5 years ago
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    \[y'=\frac{x^2 \sqrt{3x-2}}{(x+1)^2} \cdot \frac{(x+1)(4(3x-2)+3x)-12x^2+8x}{2x(3x-2)(x+1)}\] \[y'=\frac{ x^2 (3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{(x+1)(12x-8+3x)-12x^2+8x}{2x(3x-2)(x+1)}\] \[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{(x+1)(15x-8)-12x^2+8x}{2x(3x-2)(x+1)}\] \[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{15x^2-8x+15x-8-12x^2+8x}{2x(3x-2)(x+1)}\]

  14. myininaya
    • 5 years ago
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    \[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{3x^2+15x-8}{2x(3x-2)(x+1)}\]

  15. myininaya
    • 5 years ago
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    \[y'=\frac{x^2 (3x^2+15x-8)}{2x(x+1)^3(3x-2)^{1-\frac{1}{2}}}\]

  16. myininaya
    • 5 years ago
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    you can also cancel a pair of x's

  17. anonymous
    • 5 years ago
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    OHHH, there's a (3x-2) already in the denominator. Ok, I understand now

  18. myininaya
    • 5 years ago
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    \[y'=\frac{x(3x^2+15x-8)}{2(x+1)^3(3x-2)^\frac{1}{2}}\]

  19. myininaya
    • 5 years ago
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    last thing they did was multiply the numerator out and great you got it :)

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