How to find derivative of f(x)=(2x^2+1)/(2x^2-1)
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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
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We notice that we'll have to find the derivative of a fraction, so, we'll have to use the quotient rule.
(u/v)' = (u'*v - u*v')/v^2
We'll put u = 2x^2 + 1 => u' = 4x
We'll put v = 2x^2 - 1 => v' = 4x
We'll substitute u,v,u',v' in the formula above:
f'(x) = [4x*(2x^2 - 1) - (2x^2 + 1)*4x]/(2x^2 - 1)^2
We'll factorize by 4x:
f'(x) = 4x(2x^2 - 1 - 2x^2 - 1)/(2x^2 - 1)^2
We'll combine and eliminate like terms inside brackets:
f'(x) = 4x *(-2)/(2x^2 - 1)^2
f'(x) = -8x/(2x^2 - 1)^2
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