How do I find the derivative of # y=s*sqrt(1-s^2) + cos^(-1)(s)#?

1 Answer
Konstantinos Michailidis
Sep 25, 2015

It is #dy/(ds)=-(2s^2)/(sqrt(1-s^2))#

Explanation:

It is # y(s)=s*sqrt(1-s^2) + cos^(-1)(s)# hence its derivative is

#(d(y(s)))/(ds)=sqrt(1-s^2)+s*((-2s)/(2*sqrt(1-s^2)))-1/(sqrt(1-s^2))= sqrt(1-s^2)-((1+s^2)/(sqrt(1-s^2)))=(((sqrt(1-s^2))^2)-(1+s^2))/(sqrt(1-s^2))= ((1-s^2)-(1+s^2))/(sqrt(1-s^2))=-(2s^2)/(sqrt(1-s^2))#

Remarks

The derivative of #cos^(-1)(s)=arccos(s)# is #-(1/(sqrt(1-s^2)))#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
4723 views around the world
You can reuse this answer
Creative Commons License