Question

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

Answer 1

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 ✌

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