Differentiation of
Answers
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:
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)