Question

Check the nature of the following differentiable function (i) arccos(x) + ln (x+1/x) ​

Answer 1

$(i) \frac{ - 1}{ \sqrt{ {x}^{2} + 1 } } + \frac{ {x}^{2} - 1 }{x( {x}^{2} + 1 )}$

Step-by-step explanation:

$(i) \: \: \: Let, \\ y = \arccos(x) \: \: \: and \: \: \: z = ln(x + \frac{1}{x} ) \\ then, \: \: \arccos(x) + ln(x + \frac{1}{x} ) = y + z \\ Differential \: \: of \: \: the \: \: given \: \: equation \\ \boxed{\frac{dy}{dx} + \frac{dz}{dx} }\\ Now, \: \: y = \arccos(x) \\ \: \: \: \: or, \: \: \cos(y) = x \\ Differentiating \: \: with \: \: respect \: \: to \: \: x \\ \frac{d \{ \cos(y)\}}{dx} = \frac{dx}{dx} \\ - \sin(y). \frac{dy}{dx} = 1 \\ \frac{dy}{dx} = \frac{1}{ - \sin(y)} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 1}{ \sqrt{1 - { \cos}^{2} y} } \\ \: \: \: \: \: \: \: \boxed{ = \frac{ - 1}{ \sqrt{1 - {x}^{2} } } } \\again, \: \: \: z = ln(x + \frac{1}{x} ) \\ Differentiating \: \: with \: \: respect \: \: to \: \: x \\ \frac{dz}{dx} = \frac{1}{x + \frac{1}{x} } \times (1 - \frac{1}{ {x}^{2} } ) \\ = \frac{1}{ \frac{ {x}^{2} + 1 }{x} } \times \frac{ {x}^{2} - 1 }{ {x}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{x}{ {x}^{2} + 1 } \times \frac{ {x}^{2} - 1 }{ {x}^{2} } \\ \boxed{ = \frac{ {x}^{2} - 1 }{x( {x}^{2} + 1 )} } \\so, \: Differential \: \: of \: \: the \: \: given \: \: equation \\ = \frac{dy}{dx} + \frac{dz}{dx} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \boxed{ \frac{ - 1}{ \sqrt{ {x}^{2} + 1 } } + \frac{ {x}^{2} - 1 }{x( {x}^{2} + 1 )} }$