Question

Differentiation of

$y = {x}^{x}$
​

Answer 1

$\boxed{\textbf{\large{Step-by-step explanation:}}}$

we have the given function

y = x^x

first, taking log on both the sides

log y = log x^x

∴ logy =( x) x (log x )

Differentiation wrt x,

( 1/y ) x( dy / dx )

= d /dx ( (x ) x ( logx ))

we know,

if y = u.v

then , dy/dx

= u.( dv/dx )+ v.(du/dx )

therefor,

⟹ (1/y)x(dy / dx)

= ((x) x ( 1/ x)) + (( logx) x ( 1 ))

⟹ (1/y)x ( dy / dx) = ( 1 + logx)

⟹ (dy / dx ) = y ( 1 + log x)

∴ put y = x^x

( dy / dx) = x^x ( 1 + log x)

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◾Derivatives of some standard composite functions :

1] y =[ f(x) ]^n

➡ dy / dx = n[f(x)]^(n - 1)x f'(x )

2] y = sin f (x)

➡dy / dx = cos f (x) x f' (x)

3] y = √ ( f (x ) )

➡dy / dx =( 1 / 2 √[ f(x)]) x f'(x )

4] y = k

➡ dy / dx = 0 [ Note : it is not a composite function ]

5] y = cos f (x )

➡ dy /dx = -sin f (x) x f'(x )

6] y = e^([f (x) ])

➡dy /dx = e^([f (x) ]) x f'(x)

7] y = a^[f(x)]

➡ dy /dx = a^[ f (x) ] x log a x f'(x)

8] y = log [f (x) ]

➡dy /dx =( 1 / f (x) ) x [f'(x) ]

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Answer 2

Answer:

Step-by-step explanation:

Differentiation:-

Given function y=x^x

First,we take ln both sides.

The equation now:-lny=xlnx

We differentiate both sides with respect to x.

Formulae used in differentiation:-

(lnx)'=1/x(uv)'=uv'+vu'

Hence,answer is x^x(1+lnx)