Question

Differentiate y = tan-1(2 − x2)

Answer 1

Answer:

Given,

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

We know that,

$\odot \: \: \sf \: \frac{d \: {tan}^{ - 1} x}{dx} = \frac{1}{ 1 + {x}^{2} } \\ \\ \odot \: \: \sf \: \frac{d {x}^{n} }{dx} = n {x}^{n - 1} \\ \\ \odot \: \: \sf \: \frac{d \: (constant)}{dx} = 0 \\ \\ \odot \: \sf \: \frac{d \: (u \pm \: v)}{dx} = \frac{d \: u}{dx} \pm \: \frac{d \: v}{dx}$

Now

$\: \: \: \: \ \sf \: y = {tan}^{ - 1} (2 - {x}^{2} ) \\ \sf \therefore \: \frac{dy}{dx} = \frac{d \: \{ {tan}^{ - 1} (2 - {x}^{2}) \}}{dx} \\ \sf \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{1 + {(2 - {x}^{2} )}^{2} } \times \frac{d \: (2 - {x}^{2} )}{dx} \\ \sf \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{1 + 4 - 4 {x}^{2} + {x}^{4} } \times (0 - 2x) \\ \sf \therefore \: \frac{dy}{dx} = \frac{ - 2x}{ {x}^{4} - 4 {x}^{2} + 5 }$