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OpenStudy (anonymous):

use logarithmic differentiation to find y=(x^2 = 1)^x

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myininaya (myininaya):

i don't get the equation.

OpenStudy (anonymous):

y = (x^2 + 1)^x sorry about typo

myininaya (myininaya):

take ln of both sides

myininaya (myininaya):

\[\ln(y)=\ln(x^2+1)^x\]

myininaya (myininaya):

\[\ln(y)=x \ln(x^2+1)\]

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myininaya (myininaya):

take derivative of both sides: \[\frac{y'}{y}=1 \ln(x^2+1)+x \frac{2x}{x^2+1}\]

myininaya (myininaya):

not multiply both sides by y

myininaya (myininaya):

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

OpenStudy (anonymous):

wait how did you get the 1ln(x^2 +1).....

myininaya (myininaya):

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

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myininaya (myininaya):

i used product rule

myininaya (myininaya):

\[(x \ln(x^2+1))'=(x)'\ln(x^2+1)+x[\ln(x^2+1)]'\]

myininaya (myininaya):

\[1 \ln(x^2+1)+x \frac{2x}{x^2+1}\]

OpenStudy (anonymous):

okay I multiplied by y, now how can I simplify it?

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