Question

find derivative of
$tan^{ - 1} (x \ \sqrt{a}^{2} - x {}^{2} )$
differentiate with respect to x​

Answer 1

Correct Question :-

find derivative of $\rm{ \tan {}^{ - 1} \frac{x}{ \sqrt{a {}^{2} - x {}^{2} } } }$ differentiate with respect to x.

Solution :-

Let,

$\rm{y = \tan {}^{ - 1} \frac{x}{ \sqrt{a {}^{2} - x {}^{2} } } }$

$\rm{x = a \sin \theta }$

Then,

$\rm{ \therefore \theta = \sin {}^{ - 1} \frac{x}{a} \: \: \: \: \: - - - - - (1) }$

Now,

$\rm{y = \tan {}^{ - 1} \frac{a \sin \theta }{ \sqrt{a {}^{2} - a {}^{2} \sin {}^{2} \theta } } }$

It becomes,

$\rm : \longmapsto{ \tan {}^ {- 1} \frac{a \sin \theta }{ \sqrt{a {}^{2} \cos {}^{2} \theta } } }$

$\rm : \longmapsto{ \tan {}^{ - 1} \frac{a \sin \theta }{a \cos \theta } }$

$\rm : \longmapsto{ \tan {}^{ - 1} \tan \theta \: \: \bigg\{ \because \frac{ \sin \theta }{ \cos \theta } = \tan \theta \bigg\} }$

Now, it becomes

$\rm{ y = \theta}$

$\rm : \longmapsto{y = \theta = \sin {}^{ - 1} \frac{x}{a} \: \: \: \: \{by \: \: \: - (1) \}}$

Differentiate with respect to x

$\rm : \longmapsto{ \frac{d}{dx}y = \frac{d}{dx} \sin {}^{ - 1} \frac{x}{a} }$

$\rm : \longmapsto{ \frac{1}{a} \times \frac{1}{ \sqrt{1 - \frac{x {}^{2} }{a {}^{2} } } } }$

$\rm : \longmapsto{ \frac{1}{a} \times \frac{a}{ \sqrt{a {}^{2} - x {}^{2} } } = \frac{1}{ \sqrt{a {}^{2} - x {}^{2} } } }$

Hence,

${\boxed{\rm{\red{ \frac{d}{dx} \tan {}^{ - 1} \frac{x}{ \sqrt{a {}^{2} - x {}^{2} } } = \frac{1}{ \sqrt{a {}^{2} - x {}^{2} } } }}}}$

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Answer 2

Correction in the question :-

Find derivative of $\sf \fbox \red{{tan}^{ - 1} \frac{x}{ \sqrt{ {a}^{2} } - \sqrt{ {x}^{2}}}}$ Differentiate with respect to $\sf x$.

SOLUTION :-

Let,

$\sf \fbox \orange{y = {tan}^{ - 1} \frac{x}{ \sqrt{ {a}^{2} } - \sqrt{ {x}^{2}}}}$

$\sf \fbox \orange{x = a \: sin \: \theta}$

Here,

⟼ $\sf \theta = {sin}^{ - 1} \frac{x}{a}$ ---------❶

Now,

$\sf y = {tan}^{ - 1} \frac{a \: sin \: \theta}{ \sqrt{ {a}^{2} - {a}^{2} {sin}^{2} \theta}}$

On reducing it becomes,

⟼$\sf {tan}^{ - 1} \frac{a \: sin \: \theta}{ \sqrt{ {a}^{2} {cos}^{2} \theta} }$

⟼ $\sf {tan}^{ - 1} \frac{a \: sin \: \theta}{a \: cos \: \theta}$

⟼ $\sf {tan}^{ - 1}tan \: \theta \: {\therefore \: \frac{sin \: \theta}{cos \: \theta} = tan \: \theta}$

Now it becomes,

$\sf y = \theta$

⟼ $\sf y = \theta = {sin}^{ - 1} \frac{x}{a}$ by ---------❶

Differentiate with respect to $\sf x$

⟼ $\sf \frac{d}{dx}y = \frac{d}{dx} {sin}^{ - 1} \frac{x}{a}$

⟼ $\sf \frac{1}{a} \times \frac{1}{ \sqrt{1 - \frac{ {x}^{2} }{ {a}^{2} } } }$

⟼ $\sf \frac{1}{a} \times \frac{a}{ \sqrt{ {a}^{2} - {x}^{2}}} = \frac{1}{ \sqrt{ {a}^{2} - {x}^{2}}}$

Hence,

$\sf \fbox \green{\frac{d}{dx} {tan}^{ - 1} \frac{x}{ \sqrt{ {a}^{2} - {x}^{2}}} = \frac{1}{ \sqrt{ {a}^{2} - {x}^{2}}}}$

Hence Proved.

Hope it helps ✌

$\fbox \orange{ \sf @IIMrVelvetII}$

$\fbox \green{ \sf Thank You!!!}$