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

Jan 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{2}^{x} \left(x \left({x}^{2} - 1\right) \ln 2 - 3 {x}^{2} + 1\right)}{{x}^{2} {\left({x}^{2} - 1\right)}^{2}}$

Explanation:

$x y = {2}^{x} / \left({x}^{2} - 1\right)$

$y = {2}^{x} / \left(x \left({x}^{2} - 1\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x \left({x}^{2} - 1\right) \cdot {2}^{x} \ln 2 - {2}^{x} \left(3 {x}^{2} - 1\right)}{{x}^{2} {\left({x}^{2} - 1\right)}^{2}}$
(Quotient rule, Power rule and standard differential)

$= \frac{{2}^{x} \left(x \left({x}^{2} - 1\right) \ln 2 - 3 {x}^{2} + 1\right)}{{x}^{2} {\left({x}^{2} - 1\right)}^{2}}$