Question

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

Answer 1

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