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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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