Question

How do you find the derivative of `xy=(2^x)/((x^2)-1)`?

Answer 1

`dy/dx=(2^x(x(x^2-1)ln2-3x^2+1))/(x^2(x^2-1)^2)`

Explanation:

`xy=2^x/(x^2-1)`

`y=2^x/(x(x^2-1))`

`dy/dx=(x(x^2-1)*2^xln2-2^x(3x^2-1))/(x^2(x^2-1)^2)`
(Quotient rule, Power rule and standard differential)

`= (2^x(x(x^2-1)ln2-3x^2+1))/(x^2(x^2-1)^2)`