How to find result of diff of #y=xsqrt(1-x^2)-arccos x#?

2 Answers
Ratnaker Mehta
Mar 2, 2018

# dy/dx=2sqrt(1-x^2)#.

Explanation:

I hope the Question is to find

#dy/dx, if y=xsqrt(1-x^2)-arccosx#.

We have, #dy/dx=d/dx{xsqrt(1-x^2)-arccosx}#,

#=d/dx{xsqrt(1-x^2)}-d/dx{arc cosx}#,

#=xd/dx{sqrt(1-x^2)}+sqrt(1-x^2)d/dx{x}-{-1/sqrt(1-x^2)}#,

#=x{1/(2sqrt(1-x^2))}d/dx(1-x^2)+sqrt(1-x^2)+1/sqrt(1-x^2)#,

#=x/(2sqrt(1-x^2))(-2x)+sqrt(1-x^2)+1/sqrt(1-x^2)#,

#=-x^2/sqrt(1-x^2)+sqrt(1-x^2)+1/sqrt(1-x^2)#,

#={-x^2+(1-x^2)+1}/sqrt(1-x^2)#,

#={2(1-x^2)}/sqrt(1-x^2)#.

# rArr dy/dx=2sqrt(1-x^2)#.

Ratnaker Mehta
Mar 3, 2018

# dy/dx=2sqrt(1-x^2)#.

Explanation:

Here is a Second Solution to the Problem :

Note that :

#y=xsqrt(1-x^2)-arc cosx" is defined" iff |x| lt 1#.

Now, #|x|lt1rArr EE" a unique "theta in (0,pi), s.t., x=costheta#.

#x=costheta, theta in (0,pi) hArr arc cosx =theta, |x| lt 1#.

#:. y=xsqrt(1-x^2)-arc cosx#,

#rArr y=costheta*sqrt(1-cos^2theta)-theta#,

#=costhetasintheta-theta#.

#rArr y=1/2*sin2theta-theta, where, theta=arc cosx#.

Thus, #y# is a function of #theta#, and #theta" of "x#.

By the B then, we have,

#dy/dx=dy/(d theta)*(d theta)/dx#,

#=d/(d theta){1/2*sin2theta-theta}*d/dx{arc cosx}#,

#={1/2(cos2theta)d/(d theta)(2theta)-1}*{-1/sqrt(1-x^2)}#,

#=-{1/2(cos2theta)(2)-1}{1/sqrt(1-x^2)}#,

#(1-cos2theta)/sqrt(1-x^2)#,

#=(2sin^2theta)/sqrt(1-x^2)#,

#={2(1-cos^2theta)}/sqrt(1-x^2)#,

#={2(1-x^2)}/sqrt(1-x^2)#.

#rArr dy/dx=2sqrt(1-x^2)#, as in the First Solution!

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