## University Calculus: Early Transcendentals (3rd Edition)

$$y'=2^x\ln 2$$
$$y=2^x$$ Recall Theorem 5: $$\frac{d}{dx}a^u=a^u\ln a\frac{du}{dx}$$ Apply here to find $y'$: $$y'=(2^x)'=2^x\ln 2(x)'$$ $$y'=2^x\ln 2$$