Question

find dy /dx when y=tan-1 (x2)​

Answer 1

Answer:

Important Property:

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

Step-by-step explanation:

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

$let \: {x}^{2} = p$

$= > y = {tan}^{ - 1} p$

Differentiating on both sides, we get

$\frac{dy}{dx} = \frac{d}{dx} ( {tan}^{ - 1} p)$

$= > \frac{dy}{dx} = \frac{1}{1 + {p}^{2} } \times \frac{dp}{dx}$

$= > \frac{dy}{dx} = \frac{1}{1 + {( {x}^{2} })^{2} } \times \frac{d}{dx} ( {x}^{2} )$

$= > \frac{dy}{dx} = \frac{1}{1 + {x}^{4} } \times 2x$

$= > \frac{dy}{dx} = \frac{2x}{1 + {x}^{4} }$