Mathematics OpenStudy (anonymous):

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

$\ln(y)=x \ln(x^2+1)$ 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):

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