## Calculus: Early Transcendentals 8th Edition

$$k'(r)=e^r+er^{e-1}$$
$$k(r)=e^r+r^e$$ So, $$k'(r)=\frac{d}{dr}(e^r)+\frac{d}{dr}(r^e)$$ We have $\frac{d}{dr}(e^r)=e^r$ and $\frac{d}{dr}(r^e)=er^{e-1}$ Therefore, $$k'(r)=e^r+er^{e-1}$$