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Question Number 184188 by Mastermind last updated on 03/Jan/23

Differentiate, y = x^(x−1)       hi

$$\mathrm{Differentiate},\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{x}−\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{hi} \\ $$

Answered by SEKRET last updated on 03/Jan/23

      y = x^(x−1)        ln(y)  = ln(x^(x−1) )       ln(y) = (x−1)∙ ln(x)       ln(y) = x∙ln(x) − ln(x)        ( ln(y)) ′  = (x∙ln(x) − ln(x) ) ′          ((y ′)/y)  =  ln(x) + 1 − (1/x)      y′ =y∙(ln(x) + 1 − (1/x) )         y ′ = x^(x−1)  ∙(ln(x) + 1 − (1/x))       y ′ = x^(x−2) ∙(x∙ln(x) + x − 1)

$$\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{y}}\right)\:\:=\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{y}}\right)\:=\:\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\centerdot\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{y}}\right)\:=\:\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:−\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\:\:\:\:\:\:\left(\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{y}}\right)\right)\:'\:\:=\:\left(\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:−\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:\right)\:' \\ $$$$\:\:\:\:\:\:\:\:\frac{\boldsymbol{\mathrm{y}}\:'}{\boldsymbol{\mathrm{y}}}\:\:=\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:+\:\mathrm{1}\:−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{y}}'\:=\boldsymbol{\mathrm{y}}\centerdot\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:+\:\mathrm{1}\:−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:\right)\: \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\:'\:=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} \:\centerdot\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:+\:\mathrm{1}\:−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right) \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{y}}\:'\:=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}−\mathrm{2}} \centerdot\left(\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:+\:\boldsymbol{\mathrm{x}}\:−\:\mathrm{1}\right) \\ $$

Commented by Mastermind last updated on 03/Jan/23

Good!

$$\mathrm{Good}! \\ $$

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