Question

y=x^(x+1/x),Find dy/dx for the given function y wherever defined

Answer 1

Given, $\bf{y=x^{(x+1/x)}}$

or, $\bf{y=e^{(x+1/x)logx}}$

differentiate with respect to x,

$\frac{dy}{dx}=\frac{d\{e^{(x+1/x)logx}\}}{dx}$

$=e^{(x+1/x)logx}\frac{d\{(x+1/x)logx\}}{dx}$

$=x^{(x+1/x)}[logx(1-\frac{1}{x^2})+(x+1/x)\frac{1}{x}]$

hence, $\frac{dy}{dx}=x^{(x+1/x)}[logx(1-\frac{1}{x^2})+(x+1/x)\frac{1}{x}]$

Answer 2

HELLO DEAR,





GIVEN:-
Y = x^{x + 1/x}
taking log both side,

logy = (x + 1/x).logx

(1/y)*dy/dx = (x + 1/x).1/x + logx . (1 - 1/x²)

dy/dx = x^{x + 1/x}[1 + 1/x² + logx - logx/x²]



I HOPE ITS HELP YOU DEAR,
THANKS