Differentiate Inverse Cos – Proof
January 8th, 2009
How to differentiate cos-1x
y=cos-1x
Bring the cos across
cosy = x
Differentiate both sides, remember when differentiating y time by dy/dx
-sin(y) dy/dx = 1
dy/dx = -1/siny
However we want to get the differential in terms of x, to do this we can use the identity
sin2t+cos2t = 1
so
sint = √(1 – cos2t)
putting this into our expression for dy/dx we get
dy/dx = -1/√(1-cos2y)
but cosy = x so
dy/dx =- 1/√(1-x2)
Comments Left
He he he, I think you must take the following conclusions:
the differentiate of cos-1(x) is dy/dx = – 1/√(1-x2)
while
the differentiate of sin-1(x) is dy/dx = 1/√(1-x2)
How about you all?
Hmm, if you want to know an application of both cos-1(x) and sin-1(x) simultaneously, please read rohedi’s comment on
http://unapologetic.wordpress.com/2008/10/10/the-exponential-differential-equation/.
@Denaya
No no no, you are wrong Denaya, rohedi’s comment corresponds to your purpose is at this address
http://unapologetic.wordpress.com/2008/10/13/sine-and-cosine/
when Mr. Rohedi solves f”+f=0
I think you have done big blunder, huahuahau…
@Denaya
No no no, you are wrong Denaya, rohedi’s comment corresponds to your purpose is at this address
http://unapologetic.wordpress.com/2008/10/13/sine-and-cosine/
when Mr. Rohedi solves f”+f=0
I think you have done big blunder, huahuahau…sorry honey.
You’ve missed the minus!!!
d/dy(arccos) = – 1/ sq(1-x^2)
oops sorry thanks for pointing it out
dave
Oh Come on!!! I got it all the way till “dy/dx = -1/√(1-cos2y)” but then you said “but cosy = x so… dy/dx =- 1/√(1-x2)”, Not clear on how and why it is. You didn’t explain how you went from that to this dy/dx =- 1/√(1-x2)????
Thanks
@No
Look back at the start:
We are trying to differentiate the function y = cos-1(x) but this is equivalent to cosy = x (by definition of cos-1)
So then where we ave cos^2(y) in the line above it becomes x^2 by substitution