## Derivative No. 3

Using the methods in this post, I would like to evaluate $\frac{dz(r)}{dr}\bigg|_{r^+}$ with $z(r)=r\cos\theta$. Substituting, the expression to evaluate is $ \frac{d (r \cos\theta)}{dr} \bigg|_{r^+}$. From the product rule, $ \frac{dz(r)}{dr} \bigg|_{r^+} = r \frac{d( \cos\theta )}{dr}\big|_{r^+} + \cos\theta \frac{dr}{dr}\big|_{r^+}$ Each $x$ in the product rule has been replaced by $r$. Since $ \cos\theta $ does not depend on …