Question

Differentiate the function w.r.t.x. : $(1+\frac{1}{x})^{x}$

Answer 1

1\2+1 is ur answer          scmks                                                                                                                              

Answer 2

Answer:

$\frac{d}{dx}(1+\frac{1}{x}) ^{x}=(1+\frac{1}{x}) ^{x}[(\frac{-1}{1+x })+log(1+\frac{1}{x} )]$

Step-by-step explanation:

To differentiate this function we must use lograthmic differentiation

let

$y=(1+\frac{1}{x}) ^{x}$

taking log both sides

$log\:y= x\:log(1+\frac{1}{x})$

now differentiating both sides

and applying U.V. type in RHS

$(1+\frac{1}{x}) ^{x}\\ \\\frac{d}{dx} log\:y= \frac{d}{dx} [x\:log(1+\frac{1}{x} )]\\ \\ \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}\:log(1+\frac{1}{x})+log(1+\frac{1}{x} ).\frac{dx}{dx} \\ \\ \\\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{1+\frac{1}{x}})\frac{d(1+\frac{1}{x})}{dx} +log(1+\frac{1}{x} ).1 \\ \\\\ \frac{1}{y}\frac{dy}{dx}=x(\frac{x}{1+x })(0-\frac{1}{x^{2} })+log(1+\frac{1}{x} )\\ \\ \frac{1}{y}\frac{dy}{dx}=(\frac{-1}{1+x })+log(1+\frac{1}{x} )$

$\frac{dy}{dx}=y[(\frac{-1}{1+x })+log(1+\frac{1}{x} )]\\ \\ \\\frac{dy}{dx}=(1+\frac{1}{x})^{x}[(\frac{-1}{1+x })+log(1+\frac{1}{x} )]$