We have to find the derivative of f(x)=(2x^2+1)/(2x^2-1)
We use the quotient rule here, which gives the derivative of f(x)/g(x) as
(f'(x)*g(x)* - f(x)*g'(x))/(g(x))^2
f(x)=(2x^2+1)/(2x^2-1)
f'(x) = [(2x^2+1)'*(2x^2-1) - (2x^2+1)*(2x^2-1)']/(2x^2-1)^2
=> [4x*(2x^2-1) - (2x^2+1)*4x]/(2x^2-1)^2
=> [8x^2 - 4x - 8x^2 - 4x]/(2x^2-1)^2
=> -8x/(2x^2-1)^2
The required result is -8x/(2x^2-1)^2
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We have to find the derivative of f(x)=(2x^2+1)/(2x^2-1)
We use the quotient rule here, which gives the derivative of f(x)/g(x) as
(f'(x)*g(x)* - f(x)*g'(x))/(g(x))^2
f(x)=(2x^2+1)/(2x^2-1)
f'(x) = [(2x^2+1)'*(2x^2-1) - (2x^2+1)*(2x^2-1)']/(2x^2-1)^2
=> [4x*(2x^2-1) - (2x^2+1)*4x]/(2x^2-1)^2
=> [8x^2 - 4x - 8x^2 - 4x]/(2x^2-1)^2
=> -8x/(2x^2-1)^2
The required result is -8x/(2x^2-1)^2