Find $\displaystyle \frac{dy}{dx}$ by logarithmic Differentiation:
$\displaystyle y = x^2(x^2 - 1)^3(x+1)^4$
The back of the book says $\displaystyle 2x(x+1)^6(x-1)^2(6x^2 - 2x - 1)$
How?
y = (x^2)[(x^2 -1)^3][(x+1)^4]
Take the ln of both sides,
ln(y) = 2ln(x) +3ln(x^2 -1) +4ln(x+1)
Diffrentiate both sides with respect to x,
(1/y)(dy/dx) = 2/x +[3/(x^2 -1) *2x] +4/(x+1)
(1/y)(dy/dx) = 2/x +6x/(x^2 -1) +4/(x+1)
Combine the RHS into one fraction only, LCD = x(x^2 -1).
Note that x^2 -1 = (x+1)(x-1), so,
(1/y)(dy/dx) = [2(x^2 -1) +6x(x) +4x(x-1)] / [x(x^2 -1)]
(1/y)(dy/dx) = [2x^2 -2 +6x^2 +4x^2 -4x] / [x(x^2 -1)]
(1/y)(dy/dx) = [12x^2 -4x -2] / [x(x^2 -1)]
(1/y)(dy/dx) = [2(6x^2 -2x -1)] / [x(x^2 -1)]
Now to isolate dy/dx, multiply both sides by y,
dy/dx = [2(6x^2 -2x -1) / x(x^2 -1)] *{(x^2)[(x^2 -1)^3][(x+1)^4]}
dy/dx = [2(6x^2 -2x -1)] *{(x)[(x^2 -1)^2][(x+1)^4]}
dy/dx = [2x(6x^2 -2x -1)][(x^2 -1)^2][(x+1)^4]
dy/dx = [2x(6x^2 -2x -1)][((x -1)(x+1))^2][(x+1)^4]
dy/dx = [2x(6x^2 -2x -1)][(x -1)^2][(x+1)^6] --------------answer.
Or, to follow the back of the book,
dy/dx = 2x[(x+1)^6][(x-1)^2][6x^2 -2x -1] --------------answer.