1. ## logarithmic differentiation

the question is

y = (sin 2x)^3 (x^4 - 4x)^6 / (cos x) + e^(3x), find y' using log diff.

what i did so far is,

ln y = ln[(sin 2x)^3] + ln[(x^4 - 4x)^6] - @@@@@@@@@@

at @@@@@@

should it be ln[(cos x) + e^(3x)] or ln(cos x) - ln [e^(3x)] ?

2. Hello, haebinpark!

You can't split logs like that!

Find the derivative by logarithmic differen tiation:

. . $\displaystyle y \;=\; \frac{(\sin 2x)^3 (x^4 - 4x)^6}{\cos x + e^{3x}}$

Take logs: .$\displaystyle \ln(y) \;=\;\ln\left[\frac{(\sin2x)^3(x^4-4x)^6}{\cos x + e^{3x}}\right]$

. . . . . . . . $\displaystyle \ln(y) \;=\;\ln\left[(\sin2x)^3(x^4-4x)^6\right] - \ln(\cos x + e^{3x})$

. . . . . . . . $\displaystyle \ln(y)\;=\; \ln(\sin2x)^3 + \ln(x^4-4x)^6 - \ln(\cos x + e^{3x})$

. . . . . . . . $\displaystyle \ln(y) \;=\;3\ln(\sin2x) + 6\ln(x^4-4x) - \ln(\cos x + e^{3x})$

Then: .$\displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;\;=\;\;3\cdot\frac{1}{\sin2x}\cdot2\cos2x \;+\; 6\cdot\frac{1}{x^4-4x}\cdot(x^3-4) \;-\; \frac{1}{\cos x + e^{3x}}\cdot(-\sin x + 3e^{3x})$

. . . . .$\displaystyle \frac{1}{y}\!\cdot\!\frac{dy}{dx} \;\;=\;\;\frac{6\cos2x}{\sin2x} + \frac{24(x^3-1)}{x(x^3-4)} + \frac{\sin x - 3e^{3x}}{\cos x + e^{3x}}$

. . . . . . .$\displaystyle \frac{dy}{dx} \;=\;y\left[6\cot2x + \frac{24(x^3-1)}{x(x^3-4)} + \frac{\sin x + e^{3x}}{\cos x + e^{3x}} \right]$

. . . . . . .$\displaystyle \frac{dy}{dx} \;=\;\frac{(\sin2x)^3(x^4-4x)^6}{\cos x + e^{3x}}\cdot\left[6\cot2x + \frac{24(x^3-1)}{x(x^3-4)} + \frac{\sin x - 3e^{3x}}{\cos x + e^{3x}}\right]$

3. i think i would go with ln[(cos x) + e^(3x)]

y = (sin 2x)^3 (x^4 - 4x)^6 / (cos x) + e^(3x)
ln y = ln[(sin 2x)^3] + ln[(x^4 - 4x)^6] - ln[(cos x) + e^(3x)]
= 3ln(sin 2x) + 6ln(x^4 - 4x) - ln[(cos x) + e^(3x)]
(1/y)(y') = (3 cos 2x/sin 2x) + (6(4x^3 - 4)/x^4 - 4x) - [(-sin x + 3e^3x)/(cos x) + e^(3x)]

y' = [y][(3 cos 2x/sin 2x) + (6(4x^3 - 4)/x^4 - 4x) - [(-sin x + 3e^3x)/(cos x) + e^(3x)]]

y' = [(sin 2x)^3 (x^4 - 4x)^6 / (cos x) + e^(3x)][(3 cos 2x/sin 2x) + (6(4x^3 - 4)/x^4 - 4x) - [(-sin x + 3e^3x)/(cos x) + e^(3x)]]

did i do it right...?
would be highly appreciate if someone point me out if i did something wrong
thanks

4. Originally Posted by Soroban
Hello, haebinpark!

You can't split logs like that!

Take logs: .$\displaystyle \ln(y) \;=\;\ln\left[\frac{(\sin2x)^3(x^4-4x)^6}{\cos x + e^{3x}}\right]$

. . . . . . . . $\displaystyle \ln(y) \;=\;\ln\left[(\sin2x)^3(x^4-4x)^6\right] - \ln(\cos x + e^{3x})$

. . . . . . . . $\displaystyle \ln(y)\;=\; \ln(\sin2x)^3 + \ln(x^4-4x)^6 - \ln(\cos x + e^{3x})$

. . . . . . . . $\displaystyle \ln(y) \;=\;3\ln(\sin2x) + 6\ln(x^4-4x) - \ln(\cos x + e^{3x})$

Then: .$\displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;\;=\;\;3\cdot\frac{1}{\sin2x}\cdot2\cos2x \;+\; 6\cdot\frac{1}{x^4-4x}\cdot(x^3-4) \;-\; \frac{1}{\cos x + e^{3x}}\cdot(-\sin x + 3e^{3x})$

. . . . .$\displaystyle \frac{1}{y}\!\cdot\!\frac{dy}{dx} \;\;=\;\;\frac{6\cos2x}{\sin2x} + \frac{24(x^3-1)}{x(x^3-4)} + \frac{\sin x - 3e^{3x}}{\cos x + e^{3x}}$

. . . . . . .$\displaystyle \frac{dy}{dx} \;=\;y\left[6\cot2x + \frac{24(x^3-1)}{x(x^3-4)} + \frac{\sin x + e^{3x}}{\cos x + e^{3x}} \right]$

. . . . . . .$\displaystyle \frac{dy}{dx} \;=\;\frac{(\sin2x)^3(x^4-4x)^6}{\cos x + e^{3x}}\cdot\left[6\cot2x + \frac{24(x^3-1)}{x(x^3-4)} + \frac{\sin x - 3e^{3x}}{\cos x + e^{3x}}\right]$

from 4th last line,
how did it come out as 2 cos 2x ?