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

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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)

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}}

Thanks ;)

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

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)}}

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

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}}

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