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anonymous
 5 years ago
Use logarithmic differentiation...
anonymous
 5 years ago
Use logarithmic differentiation...

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y = \frac{x^{2}\sqrt{3x2}}{(x+1)^{2}} \] and x > 2/3

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2Take ln() of both sides.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2\[\ln(y)=\ln(\frac{x^2 \sqrt{3x2}}{(x+1)^2})\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2Use properties of log to expand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Currently I'm doing this: \[lny = 2lnx+\frac{1}{2}\ln(3x2)2\ln(x+1)\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2\[(\ln(f(x)))'=\frac{f'(x)}{f(x)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then I differentiate that and move the y over to get: \[\frac{dy}{dx} = y\left[\frac{3x^{2}+15x8}{2x(3x2)(x+1)} \right]\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0they give the answer to be:\[\frac{(3x^{3}+15x^{2}8x)}{2(x+1)^{3}\sqrt{3x2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0At 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
 5 years ago
Best ResponseYou've already chosen the best response.2\[y'=y(2 \frac{1}{x}+\frac{1}{2}\frac{3}{3x2}2\frac{1}{x+1}) \] \[y' = \frac{x^{2}\sqrt{3x2}}{(x+1)^{2}}(\frac{2(2)(3x2)(x+1)}{2x(3x2)(x+1)}+\frac{3x(x+1)}{2x(3x2)(x+1)}\frac{2x(2)(3x2)}{2x(3x2)(x+1)})\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2\[y'=\frac{x^2 \sqrt{3x2}}{(x+1)^2} \cdot \frac{4(3x2)(x+1)+3x(x+1)4x(3x2)}{2x(3x2)(x+1)}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2\[y'=\frac{x^2 \sqrt{3x2}}{(x+1)^2} \cdot \frac{(x+1)(4(3x2)+3x)12x^2+8x}{2x(3x2)(x+1)}\] \[y'=\frac{ x^2 (3x2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{(x+1)(12x8+3x)12x^2+8x}{2x(3x2)(x+1)}\] \[y'=\frac{x^2(3x2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{(x+1)(15x8)12x^2+8x}{2x(3x2)(x+1)}\] \[y'=\frac{x^2(3x2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{15x^28x+15x812x^2+8x}{2x(3x2)(x+1)}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2\[y'=\frac{x^2(3x2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{3x^2+15x8}{2x(3x2)(x+1)}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2\[y'=\frac{x^2 (3x^2+15x8)}{2x(x+1)^3(3x2)^{1\frac{1}{2}}}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2you can also cancel a pair of x's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OHHH, there's a (3x2) already in the denominator. Ok, I understand now

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2\[y'=\frac{x(3x^2+15x8)}{2(x+1)^3(3x2)^\frac{1}{2}}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2last thing they did was multiply the numerator out and great you got it :)
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