Question

Use logarithmic differentiation...

Answer 1

\[y = \frac{x^{2}\sqrt{3x-2}}{(x+1)^{2}} \] and x > 2/3

Answer 2

Take ln() of both sides.

Answer 3

\[\ln(y)=\ln(\frac{x^2 \sqrt{3x-2}}{(x+1)^2})\]

Answer 4

Use properties of log to expand

Answer 5

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

Answer 6

ok great

Answer 7

\[(\ln(f(x)))'=\frac{f'(x)}{f(x)}\]

Answer 8

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.

Answer 9

they give the answer to be:\[\frac{(3x^{3}+15x^{2}-8x)}{2(x+1)^{3}\sqrt{3x-2}}\]

Answer 10

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)

Answer 11

\[y'=y(2 \frac{1}{x}+\frac{1}{2}\frac{3}{3x-2}-2\frac{1}{x+1}) \]
\[y' = \frac{x^{2}\sqrt{3x-2}}{(x+1)^{2}}(\frac{2(2)(3x-2)(x+1)}{2x(3x-2)(x+1)}+\frac{3x(x+1)}{2x(3x-2)(x+1)}-\frac{2x(2)(3x-2)}{2x(3x-2)(x+1)})\]

Answer 12

\[y'=\frac{x^2 \sqrt{3x-2}}{(x+1)^2} \cdot \frac{4(3x-2)(x+1)+3x(x+1)-4x(3x-2)}{2x(3x-2)(x+1)}\]

Answer 13

\[y'=\frac{x^2 \sqrt{3x-2}}{(x+1)^2} \cdot \frac{(x+1)(4(3x-2)+3x)-12x^2+8x}{2x(3x-2)(x+1)}\]
\[y'=\frac{ x^2 (3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{(x+1)(12x-8+3x)-12x^2+8x}{2x(3x-2)(x+1)}\]
\[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{(x+1)(15x-8)-12x^2+8x}{2x(3x-2)(x+1)}\]
\[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{15x^2-8x+15x-8-12x^2+8x}{2x(3x-2)(x+1)}\]

Answer 14

\[y'=\frac{x^2(3x-2)^\frac{1}{2}}{(x+1)^2} \cdot \frac{3x^2+15x-8}{2x(3x-2)(x+1)}\]

Answer 15

\[y'=\frac{x^2 (3x^2+15x-8)}{2x(x+1)^3(3x-2)^{1-\frac{1}{2}}}\]

Answer 16

you can also cancel a pair of x's

Answer 17

OHHH, there's a (3x-2) already in the denominator. Ok, I understand now

Answer 18

\[y'=\frac{x(3x^2+15x-8)}{2(x+1)^3(3x-2)^\frac{1}{2}}\]

Answer 19

last thing they did was multiply the numerator out
and great you got it :)