Asked by SAMMY
Find the fourth derivative of lnx^3-5x^2
Answers
Answered by
oobleck
if you mean
y = ln(x^3-5x^2) then using the properties of logs,
y = ln(x^2(x-5)) = ln(x^2) + ln(x-5) = 2lnx + ln(x-5)
y' = 2/x + 1/(x-5) = (3x-10)/(x^2-5x)
y" = -(3x^2-20x+50)/(x^2-5x)^2
y<sup><sup>(3)</sup></sup> = (6x^3-60x^2+300x-500)/(x^2-5x)^3
y<sup><sup>(4)</sup></sup> = -6(3x^4-40x^3+300x^2-1000x+1250)/(x^2-5x)^4
y = ln(x^3-5x^2) then using the properties of logs,
y = ln(x^2(x-5)) = ln(x^2) + ln(x-5) = 2lnx + ln(x-5)
y' = 2/x + 1/(x-5) = (3x-10)/(x^2-5x)
y" = -(3x^2-20x+50)/(x^2-5x)^2
y<sup><sup>(3)</sup></sup> = (6x^3-60x^2+300x-500)/(x^2-5x)^3
y<sup><sup>(4)</sup></sup> = -6(3x^4-40x^3+300x^2-1000x+1250)/(x^2-5x)^4
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