Question

(1+x^2)DX/dy+2yx-6x^2=0​

Answer 1

Answer:

Step-by-step explanation:

y=sin−1(x1−x−−−−−√+x−−√1−x2−−−−−√)→(1)

Let x=sinθ

Then, cosθ=1−x2−−−−−√

Let x−−√=sinϕ

then, cosϕ=1−x−−−−−√

Putting these values in (1),

⇒y=sin−1(sinθcosϕ+cosθsinϕ)

⇒y=sin−1(sin(θ+ϕ))

⇒y=θ+ϕ

⇒y=sin−1x+sin−1x−−√

Now, differentiating both sides w.r.t. x,

⇒dydx=11−x2−−−−−√+11−x−−−−−√∗12x−−√

⇒dydx=11−x2−−−−−√+12x(1−