derivative of y=x^x

spezialize

New member
Joined
Sep 27, 2005
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26
can someone explain how to get the deriv of y=x^x.. I think you have to use log rules but I am unsure how to go about this.
 

Unco

Senior Member
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Jul 21, 2005
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1,134
\(\displaystyle y = x^x\)

Take natural logs

\(\displaystyle \ln{y} = \ln{\left(x^x\right)}\)

Log rule: \(\displaystyle \ln{\left(x^n\right)} = n\ln{x}\):

\(\displaystyle \ln{y} = x\ln{x}\)

Differentiate implicitly wrt x, apply the product rule to the RHS:

\(\displaystyle \frac{1}{y} \frac{dy}{dx} = \ln{x} + 1\)

Remember that we had \(\displaystyle y = x^x\) so we get

\(\displaystyle \frac{dy}{dx} = x^x(\ln{x} + 1)\)
 

pka

Elite Member
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Jan 29, 2005
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9,140
\(\displaystyle y = x^x\) leads to \(\displaystyle \ln (y) = (x)\ln (x)\).

Differentiate implicitly: \(\displaystyle \frac{{y'}}{y} = \ln (x) + 1\).

Solve: \(\displaystyle y' = y(\ln (x) + 1) = x^x (\ln (x) + 1)\)
 

spezialize

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Sep 27, 2005
Messages
26
thank you very much for explaining that. Makes sense.
 
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