Question

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

Answer 1

it is given that y = x^x + (x+ 1/x)^x

you should understand one thing , by which you can solve this type of question in a minute.
if any function y = f(x)^g(x) is given , you just convert it in exponential form.
e.g., y = e^{g(x).logf(x)}
now you can easily differentiate it .

here y = x^x + (x + 1/x)^x

y = e^{xlogx} + e^{xlog(x + 1/x)}

dy/dx = e^{xlogx}[x × 1/x + logx ] + e^{xlog(x + 1/x)} [x × 1/(x + 1/x) × {1 - 1/x²) + log(x + 1/x)]

= x^x(1 + logx) + (x + 1/x)^x [x²/(x² + 1) × (x² - 1)/x² + log(x + 1/x)]

= x^x(1 + logx) + (x + 1/x)^x [(x² - 1)/(x² + 1) + log(x + 1/x)]

Answer 2

HELLO DEAR,

GIVEN:-
Y = xˣ + (x+1/x)ˣ,

put p = x^x
taking log both side,
logp = xlogx

(1/p)*dp/dx = x.1/x + logx

dp/dx = x^x[1 + logx]-------( 1 )

put m = (x + 1/x)^x
taking log both side,
logm = xlog(x + 1/x)

(1/m)*dm/dx = [x.{(1 -1/x²) × (x + 1/x)} + log(x + 1/x)]

dm/dx = (x + 1/x)^x[log(x + 1/x) + (x² - 1)/(x² + 1)]-----( 2 )

adding----( 1 ) & ---( 2 )

dp/dx + dm/dx = x^x(1 + logx) + (x + 1/x)^x[log(x + 1/x) + (x² - 1)/(x² + 1)].

Also , y = p + m
=> dy/dx = dp/dx + dm/dx

hence, dy/dx = x^x(1 + logx) + (x + 1/x)^x[log(x + 1/x) + (x² - 1)/(x² + 1)].

I HOPE ITS HELP YOU DEAR,
THANKS