Question

Pls solve this example of differentiation 12th Class.... Appropriate answer will be marked BRAINLIEST....

Answer 1

To Differentiate:

$\star \: { \tan}^{ - 1} ( \frac{x}{ \sqrt{1 - {x}^{2} } } ) \: \: \: w.r.t \: \: { \sec}^{ - 1} (\frac{1}{2 {x}^{2} - 1 } )$

Let ,

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

and

$z = { \sec}^{ - 1} ( \frac{1}{2 {x}^{2} - 1 } )$

Here Put :

$x = \cos \theta$

For y

$\implies \: y = { \tan}^{ - 1} ( \frac{ \cos \theta}{ \sqrt{1 - { \cos}^{2} \theta } } ) \\ \\ \implies \: y = { \tan}^{ - 1} ( \frac{ \cos \theta}{ \sin \theta} ) \\ \\ \implies \: y = { \tan}^{ - 1} ( \cot \theta) \\ \\ \implies \: y = {tan}^{ - 1} ( \tan( \frac{\pi}{2} - \theta)) = \frac{\pi}{2} - \theta$

Now For Z

$\implies \: z = { \sec}^{ - 1} ( \frac{1}{2 { \cos}^{2} \theta - 1 } ) \\ \\ \implies \: z = { \sec}^{ - 1} ( \frac{1}{ \cos \: 2 \theta} ) \\ \\ \implies \: z = { \sec}^{ - 1} ( \sec \: 2 \theta) = 2 \theta$

we get :

$y \: = \frac{\pi}{2} - \frac{1}{2} .z$

Now Differentiate w.r.t Z :

$\implies \: \frac{dy}{dz} = \frac{d}{dz} ( \frac{\pi}{2} - \frac{1}{2} z) \\ \\ \implies \: \frac{dy}{dz} = 0 - \frac{1}{2} \\ \\ \implies \: \frac{dy}{dz} = - \frac{1}{2}$

Formula:

$\star \: \cos \: 2x = 2 \cos ^{2} x - 1 \\ \\ \star \: { \tan}^{ - 1} ( \tan \: x) = x \\ \\ \star \: { \sec}^{ - 1} ( \sec \: x) = x$

Answer 2

check the attachment........:)