Question

find the derivative of y= (2x^5 -4x^4 + 8x) / (x^2)

Here is what I have so far:
y^(1) = 6x^2 -8x + 8/x
y^(2) = 12 x - 8 + ?


How do you find the derivative of 8/x and it must have no negative exponents? thanx

Answer 1

y= (2x^5 -4x^4 + 8x) / (x^2)
y = 2x^3 - 4x^2 + 8x^-1 = 2x^3 - 4x^2 + 8/x
y' = 6x^2 - 8x - 8x^-2 = 6x^2 - 8x - 8/x^2
y" = 12x - 8 + 16x^-3 = 12x - 8 + 16/x^3
= (12x^4 - 8x^3 + 16)/x^3

or, using the quotient rule,

y= (2x^5 -4x^4 + 8x) / (x^2)
y' = [(10x^4 - 16x^3 + 8)(x^2) - (2x^5-4x^4+8x)(2x)]/(x^2)^2
= (10x^6 - 16x^5 + 8x^2 - 4x^6 + 8x^5 - 16x^2)/x^4
= (6x^4 - 8 x^3 - 8)/x^2

which agrees with the above.
You can do y" using the quotient rule again.