Asked by Anonymous
determine dy/dx: y=(x^2-3)^x
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Answered by
mathhelper
this is the only one with a certain degree of difficulty
y = (x^2 - 3)^x
take ln of both sides
ln y = x ln(x^2 - 3)
dy/dx / y = x(2x/(x^2 - 3)) + ln(x^2 - 3) , using the product rule
dy/dx = y[ 2x^2/(x^2 - 3) + ln(x^2 - 3)]
or
dy/dx = (x^2 - 3)^x [ 2x^2/(x^2 - 3) + ln(x^2 - 3)]
y = (x^2 - 3)^x
take ln of both sides
ln y = x ln(x^2 - 3)
dy/dx / y = x(2x/(x^2 - 3)) + ln(x^2 - 3) , using the product rule
dy/dx = y[ 2x^2/(x^2 - 3) + ln(x^2 - 3)]
or
dy/dx = (x^2 - 3)^x [ 2x^2/(x^2 - 3) + ln(x^2 - 3)]
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