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Mathematics Question on Continuity and differentiability

Find dydx:\text {Find} \ \frac {dy}{dx}: y=sin−1(2x1+x2)y=sin^{-1}(\frac {2x}{1+x^2})

Answer

The given relationship is y = sin-1(2x1+x2)(\frac {2x}{1+x^2})
Differentiating this relationship with respect to x, we obtain
ddx\frac {d}{dx}(sin y) = ddx\frac {d}{dx} (2x1+x2)(\frac {2x}{1+x^2})
  ⟹  \impliescos y dydx\frac {dy}{dx} = \frac {d}{dx}$$(\frac {2x}{1+x^2}) …….... (1)
The function, (2x1+x2)(\frac {2x}{1+x^2}), is of the form of uv\frac uv.
Therefore, by quotient rule, we obtain
ddx\frac {d}{dx} (2x1+x2)(\frac {2x}{1+x^2})=(1+x2) . ddx\frac {d}{dx}(2x) -2x . \frac {d}{dx}$$(\frac {1+x^2}{1+x^2})
= (1+x2).2−2x(0+2x)(1+x2)2\frac {(1+x^2) . 2-2x(0+2x)}{(1+x^2)^2}
= 2+2x2−4x2(1+x2)2\frac {2+2x^2-4x^2}{(1+x^2)^2}
= 2(1−x2)(1+x2)2\frac {2(1-x^2)}{(1+x^2)^2} ……..… (2)
Also, sin y = 2x1+x2\frac {2x}{1+x^2}
  ⟹  \impliescos y = 1−sin2y\sqrt {1-sin^2y} = 1−(2x1+x2)2\sqrt {1- (\frac {2x}{1+x^2})^2} = (1+x2)2−4x2(1+x2)2\sqrt\frac { (1+x^2)^2-4x^2}{(1+x^2)^2}
= (1−x2)2(1+x2)2\sqrt {\frac {(1-x^2)^2}{(1+x^2)^2}}
= 1−x21+x2 \frac {1-x^2}{1+x^2}
From (1), (2) and (3) we obtain
1−x21+x2\frac {1-x^2}{1+x^2} . dydx\frac {dy}{dx} = 2(1−x2)(1+x2) \frac {2(1-x^2)}{(1+x^2)}
  ⟹  \implies dydx\frac {dy}{dx} = 21+x2\frac {2}{1+x^2}