Differentiation plzzz help

the equation of a curve is y=√(1-x/1+x)
By first differentiating
1 − x/1+X
, obtain an expression for
dy/dx
in terms of x. Hence show that the
gradient of the normal to the curve at the point (x, y) is (1 + x)
√(1 − x^2).
ok i m posting this question 4th tym..people are giving me wrong answers.i read the mark scheme in it was written to use quotient rule and then chain rule..but still i couldn't get correct naswer can u all plzz help me
the correct answer is written in the question just show the question in the show that very answer form

assuming you mean this ====> y=√( (1-x) / (1+x) )

y = ( (1-x) / (1+x) )^(1/2)

y ' = (1/2) * ( (1-x) / (1+x) )^(-1/2) * [ ( (1+x) * -1 - (1-x) * 1 ) / (1+x)^2 ]

y ' = (1/2) * ( 1 / ( (1-x) / (1+x) )^(1/2) ) * [ ( (-1-x) - (1-x) ) / (1+x)^2 ]

y ' = ( 1 / 2√( (1-x) / (1+x) ) ) * [ ( -1-x - 1+x) ) / (1+x)^2 ]

y ' = [ 1 / ( 2√(1-x) / √(1+x) ) ] * [ ( -2 / (1+x)^2 ]

y ' = [ - √(1+x) / √(1-x) ] * [ ( 1 / (1+x)^2 ]

y ' = [ - 1 / √(1-x) ] * [ ( 1 / (1+x)^(3/2) ]

y ' = [ - 1 ] / [ √(1-x) * (1+x)^(3/2) ]

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gradient normal means as flipping the numerator and denominator with signs as the following:

y ' = √(1-x) * (1+x)^(3/2) =====> we can separate as 3/2 = 1 + 1/2

y ' = √(1-x) * (1+x)^(1/2) * (1+x)^1

y ' = √(1-x) * √(1+x) * (1+x)

y ' = √( (1-x) * (1+x) ) * (1+x)

y ' = √(1-x^2) * (1+x) =====> :)

welcome welcome :)

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