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use logarithmic differentiation to find y=(x^2 = 1)^x
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i don't get the equation.
y = (x^2 + 1)^x sorry about typo
take ln of both sides
\[\ln(y)=\ln(x^2+1)^x\]
\[\ln(y)=x \ln(x^2+1)\]
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take derivative of both sides: \[\frac{y'}{y}=1 \ln(x^2+1)+x \frac{2x}{x^2+1}\]
not multiply both sides by y
\[y'=y(\ln(x^2+1)+\frac{2x^2}{x^2+1})\]
wait how did you get the 1ln(x^2 +1).....
\[y'=(x^2+1)^x(\ln(x^2+1)+\frac{2x^2}{x^2+1})\]
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i used product rule
\[(x \ln(x^2+1))'=(x)'\ln(x^2+1)+x[\ln(x^2+1)]'\]
\[1 \ln(x^2+1)+x \frac{2x}{x^2+1}\]
okay I multiplied by y, now how can I simplify it?
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