IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ankitasingh

d/dx ( sin^2{cot^-1

(1-x)/(1+x)

sg33in

do u have the answer 2 dat ques... i mean 2 b sure of the method

rishikesh_anshu

For this type of problem we first simplify the equation and differentiate the simplified equation

sin^2{cot^-1

(1-x)/(1+x)}

since domain is [-1 , 1] ( u can check this ,as (1-x)/(1+x) >= 0)
we can put x = cos

so 1-x/1+x = tan²(

/2)
or

(1-x)/(1+x) = tan(

/2)

so cot^-1

(1-x)/(1+x) =cot^-1 tan(

/2) = pi/2 - tan^-1 tan(

/2)
=pi/2 -

/2

sin^2(pi/2 -

/2)=cos²(

/2)
we know form trigo cos²(

/2) = (1+cos

)/2

hence sin^2(pi/2 -

/2)=cos²(

/2) = (1+cos

)/2

but cos

= x
hence
=(1+x)/2
so d/dx( sin^2{cot^-1

(1-x)/(1+x)} = 1/2

aks1

let x=cos2a
then the fn reduces to d/dx(sin^2(cot^-1(tana))
=d/dx(sin^2(90-a))
=d/dx(cos^2(a))
= -sin(2a)
= -sin(cos^-1(x)) [since x=cos2a]
= -(1-x^2)^(1/2)

rsharma.ranchi_000

Dear Ankita,

This is in the form of d(uv)/dx.

So, this expression is given by sin^2 (x) d(cot^(-1)(

(1-x) / (1+x).

and I hope now you can solve the question. But unfortunately if you have

problem in solving this problem you can solve this for you.

Note : use differention of cot^(-1)x. Don't use another method to solve.

This question can be solved very easily by using differentiation of

(uv)dx as I have solved this question.

If you have any another doubt in mathematics you can ask me. I will solve

the question.

please rate.............

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