From d9eecca52a86329aa4857590b1d5cb87c8b81e6e Mon Sep 17 00:00:00 2001
From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Mon, 25 Mar 2019 22:04:38 +1100
Subject: [PATCH] [spec] reciprocal differentiation

---
 spec/calculus.md | 4 ++++
 1 file changed, 4 insertions(+)

diff --git a/spec/calculus.md b/spec/calculus.md
index ca7f85d..6c6685a 100644
--- 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 $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}$
-- 
2.47.1