Question

$Find\ \frac {dy}{dx}:$ $y=tan^{-1}( \frac {3x-x^3}{1-3x^2}),\ \ -\frac {1}{\sqrt 3}<x<\frac {1}{\sqrt 3}$

Answer 1

The given relationship is y = tan-1$( \frac {3x-x^3}{1-3x^2})$

y = tan-1$( \frac {3x-x^3}{1-3x^2})$

⇒tan y = $( \frac {3x-x^3}{1-3x^2})$ …...….. (1)

It is known that, tan y = $\frac {3tan\ \frac y3-tan^3\frac y3}{1-3tan^2\frac y3}$ …….... (2)

Comparing equations (1) and (2), we obtain

x = tan $\frac y3$

Differentiating this relationship with respect to x, we obtain

$\frac {d}{dx}$(x) = $\frac {d}{dx}$(tan $\frac y3$)

⇒1 = sec2$\frac y3$ . $\frac {d}{dx}$($\frac y3$)

⇒1 = sec2$\frac y3$ . $\frac 13$ . $\frac {dy}{dx}$

⇒$\frac {dy}{dx}$ = $\frac {3}{sec^2\frac y3}$ = $\frac {3}{tan^2\frac y3}$

∴$\frac {dy}{dx}$= $\frac {3}{1+x^2}$