# Logarithmic Differentiation

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.

HallsofIvy
Homework Helper
You titled this "logarithmic differentiation" but I see no logarithm.

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)

Mark44
Mentor
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) $\neq$ ln a + ln b