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[spec] derivative of [sin|cos|tan]^-1
author
Andrew Lorimer
<andrew@lorimer.id.au>
Thu, 21 Mar 2019 10:31:24 +0000
(21:31 +1100)
committer
Andrew Lorimer
<andrew@lorimer.id.au>
Thu, 21 Mar 2019 10:31:24 +0000
(21:31 +1100)
spec/calculus.md
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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
@@
$${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
| $\sin ax$ | $a\cos ax$ |
| $\cos x$ | $-\sin x$ |
| $\cos ax$ | $-a \sin ax$ |
| $\sin ax$ | $a\cos ax$ |
| $\cos x$ | $-\sin x$ |
| $\cos ax$ | $-a \sin ax$ |
+| $\tan f(x)$ | $f^2(x) \sec^2f(x)$ |
| $e^x$ | $e^x$ |
| $e^{ax}$ | $ae^{ax}$ |
| $ax^{nx}$ | $an \cdot e^{nx}$ |
| $e^x$ | $e^x$ |
| $e^{ax}$ | $ae^{ax}$ |
| $ax^{nx}$ | $an \cdot e^{nx}$ |
@@
-169,9
+170,16
@@
$${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
| $\log_e {ax}$ | $1 \over x$ |
| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
| $\log_e {ax}$ | $1 \over x$ |
| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
+| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ |
+| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
+| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |
<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
+## 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
$$y={x^{n+1} \over n+1} + c$$
## Antidifferentiation
$$y={x^{n+1} \over n+1} + c$$