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Question Number 121720 by Jamshidbek2311 last updated on 11/Nov/20

D o e s t h i s e x a m p l e w o r k t h a t w a y .

f (x) = x^{x} f^{'} (x) = x^{x} \cdot (l n x + 1)

Answered by Dwaipayan Shikari last updated on 11/Nov/20

f (x) = x^{x}

l o g (f (x)) = x l o g x

\frac{f^{'} (x)}{f (x)} = 1 + l o g x

f^{'} (x) = f (x) (1 + l o g x) = x^{x} (1 + l o g x)

Answered by ebi last updated on 11/Nov/20

Y e s .

f (x) = x^{x}

l e t y = x^{x}

l n y = l n x^{x}

l n y = x l n x

\frac{d}{d x} (l n y) = \frac{d}{d x} (x l n x)

\frac{1}{y} \cdot \frac{d y}{d x} = x \cdot \frac{1}{x} + l n x

\frac{1}{y} \cdot \frac{d y}{d x} = 1 + l n x

\frac{d y}{d x} = y (1 + l n x)

∴ f^{'} (x) = x^{x} (1 + l n x)

Commented by Jamshidbek2311 last updated on 11/Nov/20

t h a n k