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Question

Mathematics Question on Continuity and differentiability

Find dydx:Find\ \frac {dy}{dx}:
y=sin−1(2x1−x2, −12<x<12y=sin^{-1}(2x\sqrt {1-x^2},\ \frac {-1}{\sqrt 2}<x<\frac {1}{\sqrt2}

Answer

The given relationship is y=sin-1(2x1−x2(2x\sqrt {1-x^2}

y = sin-1(2x1−x2)(2x\sqrt {1-x^2})
⇒siny = (2x1−x2(2x\sqrt {1-x^2}
Differentiating this relationship with respect to x, we obtain
cos y dydx\frac {dy}{dx} = 2[x\frac {d}{dx}$$(\sqrt {1-x^2}) + (1−x2)(\sqrt {1-x^2}) dxdx\frac {dx}{dx}]
⇒\sqrt {1-sin^2y}$$\frac {dy}{dx} = 2[x2\frac x2. -2x1−x2\frac {2x}{\sqrt{1-x^2}}+1−x2\sqrt{1-x^2}]
⇒\sqrt {1-(2x\sqrt {1-x2)^2}}$$\frac {dy}{dx} = 2[−x2+1−x2√1−x2][\frac {-x^2+1-x^2}{√1-x^2}]
⇒1−4x2(1−x2)\sqrt {1-4x^2(1-x^2)} dydx\frac {dy}{dx}= 2[1−2x2√1−x2][\frac {1-2x^2}{√1-x^2}]
⇒(1−2x2)2\sqrt {(1-2x^2)^2} dydx\frac {dy}{dx}= 2[1−2x2√1−x2][\frac {1-2x^2}{√1-x^2}]
⇒(1-2x2)dydx\frac {dy}{dx} = 2[1−2x2√1−x2][\frac {1-2x^2}{√1-x^2}]
⇒dydx\frac {dy}{dx} = [2√1−x2][\frac {2}{√1-x^2}]