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What is y'' for y= (ln(5x))/x^3?

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xetaalpha2 | Student | Honors

Posted March 11, 2011 at 2:25 PM via web

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What is y'' for y= (ln(5x))/x^3?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 11, 2011 at 2:41 PM (Answer #1)

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We have y = ln 5x / x^3. We have to determine y''.

y = ln 5x / x^3 = ln 5x * x^-3

Use the product rule

y' = [ln 5x]' * x^-3 + ln 5x *[x^-3]'

=> y' = (5 / 5x)* x^-3 -3*ln 5x * x^-4

=> y' = x^-4 - 3*ln 5x * x^-4

=> x^-4( 1 - 3* ln 5x)

y'' = [x^-4]'( 1 - 3* ln 5x) + (x^-4)[1 - 3* ln 5x]'

=> y'' = -4*x^-5 * ( 1 - 3* ln 5x) + (x^-4)[- 3*5/5x]

=> y'' = -4*x^-5 * ( 1 - 3* ln 5x) -3*(x^-5)

=> y'' = x^-5 ( -4 + 12 ln 5x - 3)

=> y'' = x^-5(12*ln 5x - 7)

The required result is y'' = (12*ln 5x - 7)/ x^5

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giorgiana1976 | College Teacher | Valedictorian

Posted March 11, 2011 at 3:29 PM (Answer #2)

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To find the 2nd derivative, we'll have to determine the 1st derivative:

y' = (5*x^3/5x - 3x^2*ln 5x)/x^6

y' = (x^2 - 3x^2*ln 5x)/x^6

y' = x^2(1 - ln (5x)^3)/x^6

y' = (1 - ln (5x)^3)/x^4

The 2nd derivative:

y" = (-15x^4/5x - 4x^3(1 - ln (5x)^3)/x^8

y" = [-3x^3 - 4x^3(1 - ln (5x)^3)]/x^8

y" = [-3 - 4(1 - ln (5x)^3)]/x^5

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