<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
+Reciprocal derivatives:
+
+$${{dy \over dx} \over 1} = dx \over dy$$
+
## Differentiating $x=f(y)$
Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}$