explain how to find the derivative of |(x^2-12)(x^2+4)|?

Answer 1

y = |(x^2-12)(x^2+4)|

(x^2-12)(x^2+4)

(x^2-12) is positive if |x|>2*sqrt(3)
(x^2+4) is positive for all values of x

y = (x^2-12)(x^2+4) if |x|>2*sqrt(3)
y = -(x^2-12)(x^2+4) if |x|<2*sqrt(3)

If |x|>2*sqrt(3)
dy/dx = 2x(x^2+4) + 2x(x^2-12)
= 2x(x^2+4+x^2-12)
= 2x(2x^2 - 8)
= 4x(x^2-4)

If |x|<2*sqrt(3)
dy/dx = -2x(x^2+4) - 2x(x^2-12)
= -2x(x^2+4+x^2-12)
= -2x(2x^2 - 8)
= -4x(x^2-4)

Answer 2

I always finding expanding easier, so you have

x^4 - 8x^2 - 48, and the derivative is

4x^3 - 16x.

It's only the function which is affected by the abs value; the derivative may logically take on negative values if the function itself decreases in certain places.

Answer 3

derivative of |(x^2-12)(x^2+4)|

is 4x(x^2-4).

Answer 4

f(x) = (x² - 12) (x² + 4)
f `(x) = 2x (x² + 4) + 2x (x² - 12)
f `(x) = 2x (x² + 4 + x² - 12)
f `(x) = 2x (2x² - 8)
f `(x) = 4x (x² - 4)
f `(x) = 4x (x - 2) (x + 2)