Logarithmic Differentiation

  • Thread starter superjen
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  • #1
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y = (sin2x)3(x4-4x)6 divided by (cosx) + e3x

i came out with an answer

y' = (sin2x)3(x4-4x)6divided by (cosx) + e3x [3cot2x + 24x3-24 divided by x4-4x + Tanx + 3x


could someone tell me if im right?
i dont know if this is what im supose to do,
if you want me to write out exactly what i did i can.
 

Answers and Replies

  • #2
HallsofIvy
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You titled this "logarithmic differentiation" but I see no logarithm.
 
  • #3
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i'm supose to solve it using logartimic differentiation, so when i do it out i get

lny = 3ln(sin2x) + 6ln(x^4-4x) - ln(cosx) + 3xln(e)
 
  • #4
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i'm supose to solve it using logartimic differentiation, so when i do it out i get

lny = 3ln(sin2x) + 6ln(x^4-4x) - ln(cosx) + 3xln(e)
Your last two terms came from cos(x) + e^(3x) in the denominator. ln(a + b) [itex]\neq[/itex] ln a + ln b
 

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