Logarithm.... Too long, need assistance

**^_^Engineer_Adam^_^** · Jan 26th 2007, 03:35 AM

Find $\frac{dy}{dx}$ by logarithmic Differentiation:
$y = \frac{x^2(x-1)^2(x+2)^3}{(x-4)^5}$

Answer is $\frac{x(x-1)(x+2)^2}{(x - 4)^6} (2x^3 - 30x^2 - 6x +16)$
EDIT Wahhhh very sorry for the mistake!!!

**CaptainBlack** · Jan 26th 2007, 04:15 AM

Originally Posted by ^_^Engineer_Adam^_^

Find $\frac{dy}{dx}$ by logarithmic Differentiation:
$y = \frac{x^2(x-1)^2(x+2)^3}{(x-4)^5}$

Answer is $\frac{x(x-1)(x+2)^2}{(x - 4)^6} (2x^3 - 30x^2 - 6x +16)$
EDIT Wahhhh very sorry for the mistake!!!

As in the other problem, take logs:

$\ln(y) = 2\ln(x)+2\ln(x-1)+3\ln(x+2)-5\ln(x-4)$

so:

$\frac{1}{y}\,\frac{dy}{dx} = \frac{2}{x}+\frac{2}{x-1}+\frac{3}{x+2}-\frac{5}{x-4}$

$\frac{dy}{dx} = \frac{x^2(x-1)^2(x+2)^3}{(x-4)^5} \frac{2(x^3-15x^2-3x+8)}{x(x-1)(x+2)(x-4)}$

$\frac{dy}{dx} = \frac{x(x-1)(x+2)^2}{(x-4)^6} {2(x^3-15x^2-3x+8)}$

RonL

**^_^Engineer_Adam^_^** · Jan 26th 2007, 11:58 PM

How did you get
$2(x^3 - 15x^2 - 3x + 8)$ ?
Sorry because thats the part where i got stuck

This is where ive started:
$2(x-1)(x+2)(x-4) + 2(x)(x+2)(x-4) + 3(x)(x-1)(x-4) - 5(x)(x - 1)(x +2)$
I tried about 10 times but i still haven't got it right

**CaptainBlack** · Jan 27th 2007, 12:14 AM

Originally Posted by ^_^Engineer_Adam^_^

How did you get
$2(x^3 - 15x^2 - 3x + 8)$ ?
Sorry because thats the part where i got stuck

This is where ive started:
$2(x-1)(x+2)(x-4) + 2(x)(x+2)(x-4) + 3(x)(x-1)(x-4) - 5(x)(x - 1)(x +2)$
I tried about 10 times but i still haven't got it right

I'm tempted to say its just algebra, but I don't trust myself to get that right,
so I used machine assistance to simplify it.

But I will have a go:

$2(x-1)(x+2)(x-4)=2x^3-6x^2-12x+16$

$2(x)(x+2)(x-4)=2x^3-4x^2-16x$

$3(x)(x-1)(x-4)=3x^3-15x^2+12x$

$-5(x)(x - 1)(x +2)=-5x^3-5x^2+10x$

So adding these we get:

$2(x-1)(x+2)(x-4) + 2(x)(x+2)(x-4) + 3(x)(x-1)(x-4) - 5(x)(x - 1)(x +2)$

..... $=2x^3-30x^2-6x+16$

Which is what we wanted.

RonL

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