Question

hey guys please solve this ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=tan^{-1}\bigg(\dfrac{1}{1+6x^2}\bigg)}$

$\sf{\implies\,\dfrac{dy}{dx}=\dfrac{1}{1+\bigg(\dfrac{1}{1+6x^2}\bigg)^2}\,\,\cdot\,\,\dfrac{d}{dx}\bigg(\dfrac{1}{1+6x^2}\bigg)}$

$\sf{\implies\,\dfrac{dy}{dx}=\dfrac{(1+6x^2)^2}{(1+6x^2)^2+1}\,\,\cdot\,\,\dfrac{-12x}{(1+6x^2)^2}}$

$\sf{\implies\,\dfrac{dy}{dx}=\dfrac{-12x}{1+36x^4+12x^2+1}}$

$\sf{\implies\,\dfrac{dy}{dx}=\dfrac{-12x}{36x^4+12x^2+2}}$

$\sf{\implies\,\dfrac{dy}{dx}=\dfrac{-6x}{18x^4+6x^2+1}}$