Question

Find dy/dx if y = tan-1[(1+x2)1/2+ (1-x2)1/2 / (1+x2)1/2- (1-x2)1/2]

Answer 1

We have, y = tan –1[{√(1 + x2) – √(1 – x2)}/{√(1 + x2) + √(1 – x2)}]  

Put x2 = cos θ, then  

y = tan –1[{√(1 + cos θ) – √(1 – cos θ)}/{√(1 + cos θ) + √(1 – cos θ)}]  

= tan -1 [(√2 cos θ/2 – √2 sin θ/2)/(√2 cos θ/2 + √2 sin θ/2)]  

= tan -1 [(1 – tan θ/2)/(1 + tan θ/2)]  

= tan -1 [tan(π/4 – θ/2)]  

= π/4 – θ/2 = π/4 – 1/2 cos –1 x2.  

Differentiating w.r.t. x we get  

dy/dx = 0 – 1/2. d/dx(cos-1 x2)  

= – 1/2. – 1/√{1 – (x2)2}.d/dx(x2)  

= 1/2 .1/√(1 – x4). 2x  

= x/√(1 – x4). [Ans.]