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

$$y'=\frac{1-e}{t^e}$$
$$y=t^{1-e}$$ Recall the General Power Rule for Derivatives: $$\frac{d}{dt}t^n=nt^{n-1}$$ Apply here to find $y'$: $$y'=(t^{1-e})'=(1-e)t^{(1-e)-1}=(1-e)t^{-e}$$ $$y'=\frac{1-e}{t^e}$$