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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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

See
This Answer Now

Start your subscription to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your Subscription

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

Approved by eNotes Editorial Team