Question

Given $y={\tan }^{-1}\frac{5x}{1-6x^2},-\frac{1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}}$, then find $\frac{dy}{dx}$.

Answer 1

given that

$y={\tan }^{-1}\frac{5}{1-6x^2}$

Differentiate with respect to x

$y={\tan }^{-1}\frac{5}{1-6x^2}$

$\frac{dy}{dx}=\frac{1}{1+{\left(\frac{5}{1-6x^2}\right)}^2}\times \frac{d}{dx}\left(\frac{5}{1-6x^2}\right)$

$=\frac{1-6x^2}{{(1-6x^2)}^2+5^2}\times 5(-1)\frac{1}{{(1-6x^2)}^2}\times \frac{d}{dx}(1-6x^2)$

$=\frac{-5(1-6x^2)}{(1+36x^4-12x^2+25){(1-6x^2)}^2}\times (0-12x)$

$=\frac{60x(1-6x^2)}{(36x^4-12x^2+26){(1-6x^2)}^2}$

$=\frac{60x}{2(18x^4-6x^2+13)(1-6x^2)}$

$=\frac{30x}{(18x^4-6x^2+13)(1-6x^2)}$