Find the f'(x) if f(x)=(2-x^2)/8+x^2

Answer 1

f(x) = (2-x^2)/(8+x^2)

f'(x) = -[2-x^2)(2x) + (8+x^2)(-2x)]/(8+x^2)^2

derivative of a quotient is denominator times derivative of numerator minus numerator times derivative of denominator, all divided by denominator squared

or you could convert f(x) to eliminate the denominator

f(x) = (2-x^2)(8+x^2)^-1

then apply the product rule

derivative = 1st term times derivative of 2nd, + 2nd term times derivative of 1st

= (2-x^2)(-(8+x^2)^-2 + (8+x^2)^-1(-2x)

= -(2-x^2)/(8+x^2) -2x/(8+x^2)

= (-2+x^2-2x)/(8+x^2)

=(x^2-2x-2)/(8+x^2)