Question
use logarithmic differentiation to differentiate the function f(x)=2^x(x^2+2)^3(x^3-3)^7/(x^2+4)^1/2
Answers
Reiny
remember that log(product) = sum of logs of factors
let y = 2^x(x^2+2)^3(x^3-3)^7/(x^2+4)^1/2
take ln of both sides
ln y = ln 2^x + ln (x^2+2)^3 + ln (x^3 - 3)^7 - ln (x^2+4)^(1/2)
ln y = x (ln2) + 3 ln(x^2+2) + 7 ln(x^3-3) - (1/2) ln(x^2 + 4)
y' /y = ln2 + 6x/(x^2 + 2) + 21x^2/(x^3-3) - x/(x^2+4)
y' = dy/dx = y [ln2 + 6x/(x^2 + 2) + 21x^2/(x^3-3) - x/(x^2+4) ]
don't know what kind of simplification you need, but that would be just careful algebra
let y = 2^x(x^2+2)^3(x^3-3)^7/(x^2+4)^1/2
take ln of both sides
ln y = ln 2^x + ln (x^2+2)^3 + ln (x^3 - 3)^7 - ln (x^2+4)^(1/2)
ln y = x (ln2) + 3 ln(x^2+2) + 7 ln(x^3-3) - (1/2) ln(x^2 + 4)
y' /y = ln2 + 6x/(x^2 + 2) + 21x^2/(x^3-3) - x/(x^2+4)
y' = dy/dx = y [ln2 + 6x/(x^2 + 2) + 21x^2/(x^3-3) - x/(x^2+4) ]
don't know what kind of simplification you need, but that would be just careful algebra