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

1. anonymous

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

2. myininaya

Take ln() of both sides.

3. myininaya

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

4. myininaya

Use properties of log to expand

5. anonymous

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

6. myininaya

ok great

7. myininaya

$(\ln(f(x)))'=\frac{f'(x)}{f(x)}$

8. 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.

9. anonymous

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

10. 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)

11. myininaya

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

12. myininaya

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

13. myininaya

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

14. myininaya

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

15. myininaya

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

16. myininaya

you can also cancel a pair of x's

17. anonymous

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

18. myininaya

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

19. myininaya

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