Verified answer

AS GIVEN question READS AS

f(x) = x² - (4 / x² ) - 16

However I have a funny feeling that you have presented the question INCORRECTLY and you in fact mean:-

f (x) = (x² - 4) / (x ² - 16)

f ` (x) is then given by :-

(x ² - 16) ( 2x ) - (x ² - 4) ( 2x )

----------------------------------------

(x² - 16) ²

2x ³ - 32 x - 2x ³ + 8x

--------------------------

(x² - 16) ²

- 24 x

-----------------

(x² - 16) ²

f " (x) is then :-

(x - 16) ² (- 24 x) + 24x (2)(x² - 16)(2x)

---------------------------------------------

(x² - 16)^4

(x - 16) ² (- 24 x) + 96 x² (x² - 16)

---------------------------------------------

(x² - 16)^4

(- 24 x) (x² - 16) [ (x - 16) - 4x ]

-----------------------------------------

(x² - 16)^4

(24x) (3x + 16 )

-------------------

(x² - 16) ³

PS

Take care with presentation of questions.

I assume you meant to write

f(x) = (x^2 - 4)/(x^2 - 16)

derivative of x^2 - 4 is 2x

derivative of x^2 - 16 is 2x

Using the quotient rule

[(x^2 - 16) * 2x - (x^2 - 4)*2x]/ [(x^2 - 16)^2]

The numerator is

2x[x^2 - 16 - x^2 + 4] = 2x [-12] = -24x

The first derivative is

-24x/[(x^2 - 16)^2]

The derivative of -24x is -24

The derivative of (x^2 - 16)^2 is 2(x^2 - 16) * 2x = 2x(x^2 - 16)

Using the quotient rule again for the second derivative

[(x^2 - 16)^2 * (-24) - (-24x)*2x(x^2 - 16)]/[(x^2 - 16)^4]

yep you're right about the quotient rule. square the bottom, giving you (x^2-16)^2 in the denominator.

now take the derivative of the numerator and times it by the denominator

2x(x^2-16)

2x^3-32x

now the derivative of denom and times it by the numerator

2x(x^2-4)

2x^3-8x

2x^3-32x - 2x^3-8x

-24x over (x^2-16)^2

Using Liebniz's notation:

y = u/v

dy/dx = (vu'-uv')/v^2

u = x^2 - 4

u' = 2x

v = x^2 - 16

v' = 2x

v^2 = x^4 - 32x^2 + 256

dy/dx

= [(x^2 - 16).2x - (x^2 - 4).2x]/[x^4 - 32x^2 + 256]

= [-32x + 8x]//[x^4 - 32x^2 + 256]

= [-24x]/[x^4 - 32x^2 + 256]

This is probably as simplified as you would want it.