[spec] derivative of [sin|cos|tan]^-1

diff --git a/spec/calculus.md b/spec/calculus.md
index 3dbee28e75da1da92beea859687c16b536747287..ca7f85dd9993e6cf5903c8ffb56721b3fe477312 100644 (file)
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -162,6 +162,7 @@ $${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
 | $\sin ax$ | $a\cos ax$ |
 | $\cos x$ | $-\sin x$ |
 | $\cos ax$ | $-a \sin ax$ |
+| $\tan f(x)$ | $f^2(x) \sec^2f(x)$ |
 | $e^x$ | $e^x$ |
 | $e^{ax}$ | $ae^{ax}$ |
 | $ax^{nx}$ | $an \cdot e^{nx}$ |
@@ -169,9 +170,16 @@ $${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
 | $\log_e {ax}$ | $1 \over x$ |
 | $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
 | $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
+| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ |
+| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
+| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |
 
 <!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
 
+## Differentiating $x=f(y)$
+
+Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}$
+
 ## Antidifferentiation
 
 $$y={x^{n+1} \over n+1} + c$$