# How to find derivative of f(x)=(2x^2+1)/(2x^2-1)

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### 2 Answers

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**

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**