Question

${ \tan}^{ - 1} \frac{ \sqrt{x} - x }{1 + {x}^{ \frac{3}{2} } }$
differentiation .​

Answer 1

$\huge \mathfrak{solution}$

Let

$y \: = {tan}^{ - 1} \frac{ \sqrt{x} - 1 }{1 + {x}^{ \frac{3}{2} } } \\ \\ \rightarrow \: y = {tan}^{ - 1} \frac{ \sqrt{x} - 1}{1 + \sqrt{x} .x} \\ \\ \rightarrow \: y = {tan}^{ - 1} \sqrt{x} - {tan}^{ - 1} x$

Now , Differentiate w.r.t. x we get ,

$\frac{dy}{dx} = \frac{d}{dx} ( {tan}^{ - 1} \sqrt{x} - {tan}^{ - 1} x) \\ \\ \implies \: \frac{dy}{dx} = \frac{1}{1 + x} . \frac{1}{2 \sqrt{x} } - \frac{1}{1 + {x}^{2} }$

Formula used :

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

$\bold{ \frac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }$

Answer 2

Answer:

$\large\boxed{\sf{ \frac{1}{2 \sqrt{x} (1 + x)} - \frac{1}{1 + {x}^{2} } }}$

Step-by-step explanation:

To differentiate the given expression i.e.,

$\large{{ \tan}^{ - 1} \dfrac{ \sqrt{x} - x }{1 + {x}^{ \frac{3}{2} } } }$

Let's assume that,

$\sf{y = { \tan }^{ - 1} \frac{ \sqrt{x} - 1}{1 + {x}^{ \frac{3}{2} } } } \\ \\ \sf{ = > y = { \tan }^{ - 1} \frac{ \sqrt{x} - 1 }{1 + x\sqrt{x} }} \\ \\ \sf{ = > y = { \tan }^{ - 1} \sqrt{x} - { \tan }^{ - 1} x}$

Differentiating both the sides wrt x,

$\sf{= > \dfrac{dy}{dx} = \dfrac{d}{dx} ( { \tan}^{ - 1} \sqrt{x} - { \tan }^{ - 1} x) }\\ \\ \sf{= > \frac{dy}{dx} = (\frac{1}{1 + ({ \sqrt{x}) }^{2} } \times \frac{1}{2 \sqrt{x} } ) - ( \frac{1}{1 + {x}^{2} } \times 1)} \\ \\ \sf{ = > \frac{dy}{dx} = \frac{1}{2 \sqrt{x} (1 + x)} - \frac{1}{1 + {x}^{2} } }$

Concept Map :-

• $\sf{\dfrac{d}{dx} { \tan }^{ - 1} x = \dfrac{1}{1 + {x}^{2} } }$

• $\sf{\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1} }$