If f(x) = x^2 - 7 / 7-x^2, then dy/dx =

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f(x) = (x^2 - 7)/(7 - x^2) = (x^2 - 7)/-(x^2 - 7) = -1

f '(x) = 0

f(x) = (x² - 7) / (7 - x²)

u = x² - 7
u' = 2x

v = 7 - x²
v' = - 2x

f '(x) = (u'v - uv') / v²..............(Quotient Rule)

f '(x) = [2x(7 - x²) - (x² - 7)(- 2x)] / (7 - x²)²
f '(x) = [(14x - 2x³) - (- 2x³ + 14x)] / (7 - x²)²
f '(x) = (14x - 2x³ + 2x³ - 14x) / (7 - x²)²
f '(x) = 0 / (7 - x²)²
f '(x) = 0
¯¯¯¯¯¯¯
...

.........x^2 - 7
f(x) =--------------
..........7 - x^2

.........x^2 - 7
f(x) =- ------------
..........x^2 - 7

f(x) = - 1

f'(x) = 0 answer//

Sub out a negative one in the denominator. F(x) = -1, so the derivative (dy/dx) is 0.

f(x)=(x²-7)/(7-x²)
f'(x)=[2x(7-x²)-(x²-7)(-2x)]/(7-x²)²
f'(x)=0

Dy/dx = 2x - 0 - 2x = 0