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susus · Nov 7, 2010, 03:42 PM

What is the derivative of f(x) = arccos(x)

?

I reached the point of this :
I got :
f ' (x) = 1/ -sinf(x)
I know the answer supposed to be -1/sqr(1-x^2)...
but how can I get that from this point : f ' (x) = 1/ -sinf(x)

Unknown008 · Nov 7, 2010, 10:39 PM

I think you can do like this:

Let y = arccos x

Then;

$cos(y) = x$

by implicit differentiation;

$-sin(y) \frac{dy}{dx} = 1$

$\frac{dy}{dx} = -\frac{1}{sin(y)}$

You know that cos y = x.

Draw a triangle with and angle y, adjacent side x and hypotenuse 1.

The opposite side is then $\sqrt{1 - x^2}$ using Pythagoras' Theorem.

Then, we get:

$sin(y) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}$

Substitute that in your derivative:

$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}$

Voilà. :)

This means:

$f'(x) = -\frac{1}{\sqrt{1 - x^2}}$

susus · Nov 8, 2010, 06:39 AM

Thanks ;)

susus · Nov 14, 2010, 12:11 PM

what about the derivative of f(x) = arccos(x).. it did not work out with me :( haha..

Unknown008 · Nov 14, 2010, 10:36 PM

I don't understand... I gave you the derivative of f(x)...

Or if you want, you can simplify:

$f'(x) = -\frac{1}{\sqrt{1-x^2}} = \frac{1}{\sqrt{(1+x)(1-x)}}$

susus · Nov 14, 2010, 11:03 PM

sorry sorry sorry , I meant f(x)= arccotx
!!

ebaines · Nov 15, 2010, 07:53 AM

Consider a right trangle with angle $\theta$ such that $\cot(\theta) = x$ , and $Arccot(x) = \theta$ as shown in the figure.

Take the derivative of $x\ = \ \cot(\theta)$ :

$ \frac {dx} {d \theta} = \frac {-1} {sin^2 \theta} = \frac {-1} {( \frac 1 {1 + x^2} ) } = -(1+x^2) $

Rearrange:

$ \frac {d \theta} {dx} = \frac {-1} {1+x^2} = \frac {d \ Arccot(x)} {dx} $

So the derivative of Arccot(x) is:
$\frac {-1} {1 + x^2}$

Note that this is the negative of the derivative of Arctan(x). If you go through all the trig functions you'll find that the derivative of a "Arc-co" function is the always the negative of its partner. Hence:

$ \frac {d\ Arcsin(x)} {dx} = \frac 1 {\sqrt{1-x^2}} \\ \frac {d\ Arccos(x)} {dx} = \frac {-1} {\sqrt{1-x^2}} \\ \frac {d\ Arctan(x)} {dx} = \frac 1 {1+x^2}\\ \frac {d\ Arccot(x)} {dx} = \frac {-1} {1+x^2}\\ \frac {d\ Arcsec(x)} {dx} = \frac 1 {x \sqrt {x^2-1}}\\ \frac {d\ Arccsc(x)} {dx} = \frac {-1} {x \sqrt{x^2-1}} $

Unknown008 · Nov 15, 2010, 09:21 AM

If you are unsure of what $\frac{d}{dx}cot(x)$ , use the quotient rule after converting it to cos and sin.