Derivative of sin^-1 x using First Principles

Asked Dec 19 '15 at 02:37

Active Dec 19 '15 at 06:57

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I found this explanation in my math textbook and haven't found this method in any online tutorials. Here are the first few lines of finding the derivative of $\sin^{-1} (x)$ using First Principles:

Let $y = \sin^{-1} (x)$. Then, $x = \sin (y)$ and so $x + \delta x = \sin (y + \delta y)$

As $\delta x$ -> $0$, $\delta y$ -> $0$

Now, $\delta x = \sin (y + \delta y) - \sin y$

∴ $1 = {\sin(y + \delta y) - \sin y}/\delta x$

or, $1 = { \sin(y + \delta y) - \sin y / \delta x }$ $\delta y/\delta x$

and the limits on both sides are taken and lived happily ever after.

My Question: In the last step, where does the $\delta y/\delta x$ come from?

Any help is greatly appreciated.

edited Dec 19 '15 at 02:50

Martin Argerami

205,756

asked Dec 19 '15 at 02:37

blenderender

1

You can choose $\delta x$ so that there is no $\delta y$ making the equation $x+\delta x=\sin(y+\delta y)$ hold. – Matt Samuel Dec 19 '15 at 02:43
I don't get it. How would you go about doing that? – blenderender Dec 19 '15 at 02:44
1

i think the last line should read correctly: $1 = ( \frac{sin(y + \delta y)-sin(y)}{\delta y} ) \frac{\delta y}{\delta x}$ – math635 Dec 19 '15 at 02:45
@math635 Probably. This book has too many typos. How would the δy denominator help? – blenderender Dec 19 '15 at 02:54
@labbhattacharjee That's not exactly my question. I'm just confused as to how the dy/dx appeared. – blenderender Dec 19 '15 at 04:35

Derivative of sin^-1 x using First Principles

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