## Calculus (3rd Edition)

$$y'=(2^x\ln 2+3^{-x}\ln 3)x^{-1} - (2^x-3^{-x} )x^{-2}.$$
Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(a^x)'=a^x\ln{a}$ Since $y=\frac{2^x-3^{-x}}{x}$, it is easier to simplify $y$ as follows $$y=(2^x-3^{-x})x^{-1}$$ then, by the product rule, we have $$y'=(2^x\ln 2+3^{-x}\ln 3)x^{-1} - (2^x-3^{-x} )x^{-2}.$$