[spec] reciprocal differentiation

diff --git a/spec/calculus.md b/spec/calculus.md
index ca7f85dd9993e6cf5903c8ffb56721b3fe477312..6c6685a3690062312336d71d7dcce93374b83ae8 100644 (file)
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -176,6 +176,10 @@ $${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
 
 <!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
 
+Reciprocal derivatives:
+
+$${{dy \over dx} \over 1} = dx \over dy$$
+
 ## Differentiating $x=f(y)$
 
 Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}$