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Mathematics
OpenStudy (anonymous):

use logarithmic differentiation to find y = x^(lnx)

myininaya (myininaya):

\[\ln(y)=\ln(x^{\ln(x)})\]

OpenStudy (anonymous):

you have a choice. myininaya will do it one way, i will do it another

myininaya (myininaya):

\[\ln(y)=\ln(x) \ln(x)\]

myininaya (myininaya):

\[\ln(y)=[\ln(x)]^2\]

myininaya (myininaya):

\[\frac{y'}{y}=2 \ln(x) \frac{1}{x}\]

myininaya (myininaya):

now multiply y on both sides

OpenStudy (anonymous):

myininya did u do product rule?

myininaya (myininaya):

chain rule

OpenStudy (anonymous):

\[x^{\ln(x)}=e^{\ln(x)\times \ln(x)}\]so \[\frac{d}{dx}x^{\ln(x)}=\frac{d}{dx}e^{\ln(x)\times \ln(x)}\] \[=e^{\ln(x)\times \ln(x)}\times \frac{d}{dx}\ln(x)\times \ln(x)\] and \[\frac{d}{dx}\ln^2(x)=2\ln(x)\frac{1}{x}\] so your final answer is \[=e^{\ln(x)\times \ln(x)}\times \frac{2\ln(x)}{x}\]

OpenStudy (anonymous):

why did u multiply by 1/x?

OpenStudy (anonymous):

chain rule. the derivative of \[f^2(x)\] is \[2f(x)f'(x)\]

OpenStudy (anonymous):

\[\frac{d}{dx}\ln^2(x)=2\ln(x)\ln'(x)=2\ln(x)\frac{1}{x}\]

OpenStudy (anonymous):

thamk you!

OpenStudy (anonymous):

when you see \[f(x)^{g(x)}\] and you want the derivative, you have a choice. you can 1) take the log 2) take the derivative of the log 3) multiply the result by the original function

OpenStudy (anonymous):

or, you can rewrite \[f(x)^{g(x)}=e^{g(x)\ln(f(x))}\] and take the derivative using the chain rule

OpenStudy (anonymous):

it amounts to the same thing because all the work is finding the derivative of \[g(x)\ln(f(x))\]

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