Question

Differentiate
cos^-1{1-x^2\1+x^2}​

Answer 1

To differentiate :

$\star \: { \cos}^{ - 1} ( \frac{1 - {x}^{2} }{1 + {x}^{2} } )$

Here ,

• Put ,x = tan @

we get ,

$\implies \: { \cos}^{ - 1} ( \frac{1 - { \tan}^{2} \alpha }{1 + { \tan}^{2} \: \: \alpha } )$

As we know that :

$\star \: \boxed{ \cos \: 2 \theta \: = \frac{1 - { \tan}^{2} \theta}{1 + { \tan}^{2} \theta }}$

$\implies \: { \cos}^{ - 1} ( \cos2 \alpha ) \\ \\ \implies \: 2 \alpha$

Now put the value of @ .

$\implies \: 2 \: { \tan}^{ - 1} x$

Now , differentiate w.r.t x ,

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

Formula used:

$\leadsto \bold{ { \cos}^{ - 1} ( \cos \theta) = \theta}$

$\leadsto \: \bold{\frac{d}{dx} { \tan}^{ - 1} x = \frac{1}{1 + {x}^{2} } }$

Answer 2

Given ,

The function is $\tt y = {cos}^{ - 1} \{ \frac{1 - {(x)}^{2} }{1 + {(x)}^{2} } \}$

Let , x = tan(θ) => tan^(-1)x = θ

Thus ,

$\tt y = {cos}^{ - 1} \{ \frac{1 - {tan}^{2} \theta}{1 + {tan}^{2} \theta } \}$

$\tt y = {cos}^{ - 1} \{ cos2 \theta\}$

$\tt y = 2 \theta$

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

Now , differentiating y wrt x , we get

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

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

$\tt \frac{dy}{dx} = \frac{2}{1 + {(x)}^{2} }$

Remmember :

★ $\tt cos(2x) = \frac{1 - {(x)}^{2} }{1 + {(x)}^{2} }$

★ $\tt {cos}^{ - 1}\{cos(x) \} = x$

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

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