View Full Version : What is the derivative of f(x) = arccos(x)
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}}
Voilą. :)
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.