Question

differentiate with respect to x .f(x) = log xX​

Answer 1

Let y=(logx)

x

+x

logx

Also, let u=(logx)

x

and v=x

logx

∴y=u+v

⇒

dx

dy

=

dx

du

+

dx

dv

... (1)

u=(logx)

x

⇒logu=log[(logx)

x

]

⇒logu=xlog(logx)

Differentiating both sides with respect to x, we obtain

u

1

dx

du

=log(logx)+x⋅

logx

1

⋅

x

1

⇒

dx

du

=(logx)

x

[log(logx)+

logx

x

.

x

1

]

⇒

dx

du

=(logx)

x

[log(logx)+

logx

1

]

⇒

dx

du

=(logx)

x

[

logx

log(logx).logx+1

]

⇒

dx

du

=(logx)

x−1

[1+logx.log(logx)] ...(2)

v=x

logx

⇒logv=log(x

logx

)

⇒logv=logx.logx=(logx)

2

Differentiating both sides with respect to x, we obtain

v

1

dx

dv

=

dx

d

[(logx)

2

]

⇒

v

1

.

dx

dv

=2(logx).

dx

d

(logx)

⇒

dx

dv

=2v(logx).

x

1

⇒

dx

dv

=2x

logx

x

logx

⇒

dx

dv

=2x

logx−1

.logx ...(3)

Therefore, from (1), (2), and (3), we obtain

⇒

dx

dy

=(logx)

x−1

[1+logx.log(logx)]

Hope it's help to you buddy and Mark this answer as brainlist please

Answer 2

let y = log x^x

=> y = x log x [ log x^m = m log x b]

=> differentiating wrt x, we get

=> dy/dx = x * 1/x + log x * 1 [ product rule ]

[ d(uv)/dx => u * dv/dx + v * du/dx ]

=> dy/dx = 1 + log x