From 358fda608850a5882d881daeff1150a450b4254b Mon Sep 17 00:00:00 2001 From: Andrew Lorimer Date: Thu, 21 Mar 2019 21:31:24 +1100 Subject: [PATCH] [spec] derivative of [sin|cos|tan]^-1 --- spec/calculus.md | 8 ++++++++ 1 file changed, 8 insertions(+) diff --git a/spec/calculus.md b/spec/calculus.md index 3dbee28..ca7f85d 100644 --- 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 @@ $${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$ | $\log_e {ax}$ | $1 \over x$ | | $\log_e f(x)$ | $f^\prime (x) \over f(x)$ | | $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ | +| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ | +| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ | +| $\tan^{-1} x$ | $1 \over {1 + x^2}$ | +## 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 $$y={x^{n+1} \over n+1} + c$$ -- 2.43.2