Question

can some one solve this please..u need to find dy/dx for this function​

Answer 1

Answer:

-1/√(1-x)(1+x)³

Step-by-step explanation:

let y=√(1-x)/(1+x)-------------------(1)

squaring both sides

y²=1-x/1+x

let z=y²

dz/dy=2y

so dy/dz=1/2y-----------------(2)

Now z=1-x/1+x

dz/dx=1/(1+x)² [ (1+x)*d/dx(1-x)-(1-x)*d/dx(1+x) ]

=1/(1+x)²[(1+x)*(-1)-(1-x)*(1) ]

=1/(1+x)²(-1-x-1+x)

dz/dx = -2/(1+x)²-------------------(3)

We know

dy/dx=dy/dz*dz/dx

putting value from (1) and (2)

dy/dx=1/2y*(-2)/(1+x)²

=-2*1/2y/(1+x)²

=-1/y*(1+x)²

=-1/(1+x)²*√(1+x)/(1-x)

=-1/√(1-x)(1+x)³