Question

$Find\ \frac {dy}{dx}:$
$y=cos^{-1}(\frac {2x}{1+x^2}),\ -1<x<1$

Answer 1

The given relationship is y = cos-1$(\frac {2x}{1+x^2})$

y = cos-1$(\frac {2x}{1+x^2})$
⇒cosy = $\frac {2x}{1+x^2}$
Differentiating this relationship with respect to x,we obtain

$\frac {d}{dx}$(cos y) = \frac {d}{dx}$$(\frac {2x}{1+x^2})
⇒-sin y $\frac {dy}{dx}$ = $\frac {(1+x^2).\frac {d}{dx}(2x)-2x.\frac {d}{dx}(1+x^2)}{(1+x^2)^2}$
⇒-$\sqrt{1-cos^2y}$ $\frac {dy}{dx}$ = $\frac {(1+x^2).2-2x.2x}{(1+x^2)^2}$
⇒$[\sqrt {1-(\frac {2x}{1+x^2})^2}$ $\frac {dy}{dx}$ = -$[\frac {2(1-x^2)}{(1+x^2)^2}]$
⇒$\sqrt {\frac {(1-x^2)^2-4x^2}{(1+x^2)^2}}$ $\frac {dy}{dx}$ = -$\frac {2(1-x^2)}{(1+x^2)^2}$
⇒$\sqrt {\frac {(1-x^2)^2}{(1+x^2)^2}}$ $\frac {dy}{dx}$ = -$\frac {2(1-x^2)}{(1+x^2)^2}$
⇒$\frac {1-x^2}{1+x^2}$.$\frac {dy}{dx}$= -$\frac {2(1-x^2)}{(1+x^2)^2}$
⇒$\frac {dy}{dx}$ = - $\frac {2}{1+x^2}$