Differentiation of x^x......plz give the right answer...it's very urgent...give step by step solution
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Hey!!!....Here is ur answer
let x^x=y
taking log of both sides
logy=logx^x
now differentiate both sides
=> d/dx (logy)= d/dx (x logx)
=> 1/y dy/dx=x d/dx(logx)+logx d/dx (x)
=> dy/dx= y [x/x+logx]
=> dy/dx=x^x (1+logx)
Hope it will help you
let x^x=y
taking log of both sides
logy=logx^x
now differentiate both sides
=> d/dx (logy)= d/dx (x logx)
=> 1/y dy/dx=x d/dx(logx)+logx d/dx (x)
=> dy/dx= y [x/x+logx]
=> dy/dx=x^x (1+logx)
Hope it will help you
Anonymous:
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It helps a lot
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Thanks dude : )
I'm always available for ur help
thanks again bhai
: )
Answered by
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HEY BUDDY...!!!!
HERE'S THE ANSWER...
_______________________
▶️ Refer the above attachment..
HOPE HELPED..
HAVE A NICE DAY..
:-)
HERE'S THE ANSWER...
_______________________
▶️ Refer the above attachment..
HOPE HELPED..
HAVE A NICE DAY..
:-)
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