anonymous
  • anonymous
Use logarithmic differentiation...
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[y = \frac{x^{2}\sqrt{3x-2}}{(x+1)^{2}} \] and x > 2/3
myininaya
  • myininaya
Take ln() of both sides.
myininaya
  • myininaya
\[\ln(y)=\ln(\frac{x^2 \sqrt{3x-2}}{(x+1)^2})\]

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myininaya
  • myininaya
Use properties of log to expand
anonymous
  • anonymous
Currently I'm doing this: \[lny = 2lnx+\frac{1}{2}\ln(3x-2)-2\ln(x+1)\]
myininaya
  • myininaya
ok great
myininaya
  • myininaya
\[(\ln(f(x)))'=\frac{f'(x)}{f(x)}\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
they give the answer to be:\[\frac{(3x^{3}+15x^{2}-8x)}{2(x+1)^{3}\sqrt{3x-2}}\]
anonymous
  • anonymous
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)
myininaya
  • myininaya
\[y'=y(2 \frac{1}{x}+\frac{1}{2}\frac{3}{3x-2}-2\frac{1}{x+1}) \] \[y' = \frac{x^{2}\sqrt{3x-2}}{(x+1)^{2}}(\frac{2(2)(3x-2)(x+1)}{2x(3x-2)(x+1)}+\frac{3x(x+1)}{2x(3x-2)(x+1)}-\frac{2x(2)(3x-2)}{2x(3x-2)(x+1)})\]
myininaya
  • myininaya
\[y'=\frac{x^2 \sqrt{3x-2}}{(x+1)^2} \cdot \frac{4(3x-2)(x+1)+3x(x+1)-4x(3x-2)}{2x(3x-2)(x+1)}\]
myininaya
  • myininaya
\[y'=\frac{x^2 \sqrt{3x-2}}{(x+1)^2} \cdot \frac{(x+1)(4(3x-2)+3x)-12x^2+8x}{2x(3x-2)(x+1)}\] \[y'=\frac{ x^2 (3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{(x+1)(12x-8+3x)-12x^2+8x}{2x(3x-2)(x+1)}\] \[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{(x+1)(15x-8)-12x^2+8x}{2x(3x-2)(x+1)}\] \[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{15x^2-8x+15x-8-12x^2+8x}{2x(3x-2)(x+1)}\]
myininaya
  • myininaya
\[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{3x^2+15x-8}{2x(3x-2)(x+1)}\]
myininaya
  • myininaya
\[y'=\frac{x^2 (3x^2+15x-8)}{2x(x+1)^3(3x-2)^{1-\frac{1}{2}}}\]
myininaya
  • myininaya
you can also cancel a pair of x's
anonymous
  • anonymous
OHHH, there's a (3x-2) already in the denominator. Ok, I understand now
myininaya
  • myininaya
\[y'=\frac{x(3x^2+15x-8)}{2(x+1)^3(3x-2)^\frac{1}{2}}\]
myininaya
  • myininaya
last thing they did was multiply the numerator out and great you got it :)

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