Question

Find $\frac {dy}{dx}:$ $y=cos^{-1}(\frac {1-x^2}{1+x^2}),\ \ 0<x<1$

Answer 1

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

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

⇒cos y = $(\frac {1-x^2}{1+x^2})$

⇒$\frac {1-tan^2\frac y2}{1+tan^2\frac y2}$ = $\frac {1-x^2}{1+x^2}$
On comparing L.H.S. and R.H.S. of the above relationship, we obtain
tan $\frac y2$ = x
Differentiating this relationship with respect to x,we obtain

sec2$\frac y2$ . \frac {d}{dx}$$(\frac y2) = $\frac {d}{dx}$(x)

⇒sec2$\frac y2$ . \frac 12$$\frac {dy}{dx} = 1

⇒$\frac {dy}{dx}$ = $\frac {2}{sec^2\frac y2}$

⇒$\frac {dy}{dx}$= $\frac {2}{tan^2\frac y2}$

∴$\frac {dy}{dx}$ = $\frac {1}{1+x^2}$