Question

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

Answer 1

i don't get the equation.

Answer 2

y = (x^2 + 1)^x

sorry about typo

Answer 3

take ln of both sides

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

not multiply both sides by y

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

i used product rule

Answer 12

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

Answer 13

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

Answer 14

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