Sum

if `y=x^x` find `(dy)/(dx)`

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#### Solution

`y=x^x`

Taking logarithms of both sides

`logy=logx^x`

`logy=xlogx`

Differentiating both sides with respect to 'x', we get

`(1/y) dy/dx=x(1/x)+logx(1)`

`dy/dx=y(1+logx)`

`dy/dx=x^x(1+logx)`

Concept: Derivative - Exponential and Log

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