Use of logrithmic differentiation

[paul davo]

Hi I'm currently stuck doing some difficult differentiation. See as follows, I have not started this at all......

Use logarithmic differentiation to differentiate

y = (x+1) (x-2)^3
(x-3)

Any help would be much appreicated.

Best regards

Paul

 

[srmichael]

Hi I'm currently stuck doing some difficult differentiation. See as follows, I have not started this at all......

Use logarithmic differentiation to differentiate

y = (x+1) (x-2)^3
(x-3)

Any help would be much appreicated.

Best regards

Paul

First, you need to know your log rules. Do these look familiar?

log(uv) = log(u) + log(v)

log (u/v) = log(u) - log(v)

log u^v = vlog(u)

Use these rules to on your function to expand.

 

[Bob Brown MSEE]

You may find the a uTube Video useful.

 

[soroban]

Hello, paul davo!

Hi, I'm currently stuck doing some difficult differentiation.
See as follows, I have not started this at all . . .
It sounds like you were never taught how to differentiate logarithmically . . . How come?

Use logarithmic differentiation to differentiate: .$\displaystyle y \:=\: \dfrac{(x+1)(x-2)^3}{x-3}$

Take logs: .$\displaystyle \ln(y) \;=\;\ln\dfrac{(x+1)(x-2)^3}{x-3} \;=\;\ln(x+1) + \ln(x-2)^3 - \ln(x-3)$

. . . . . . . . $\displaystyle \ln(y) \;=\;\ln(x+1) + 3\ln(x-2) - \ln(x-3)$


Differentiate implicitly:

. . $\displaystyle \dfrac{1}{y}\cdot y' \;=\;\dfrac{1}{x+1} + \dfrac{3}{x-2} - \dfrac{1}{x-3}$

n . . . $\displaystyle y' \;=\;y\left(\dfrac{1}{x+1} + \dfrac{3}{x-2} - \dfrac{1}{x-3}\right)$

n . . . $\displaystyle \displaystyle y' \;=\;\frac{(x+1)(x-2)^3}{x-3}\left(\frac{1}{x+1} + \frac{3}{x-2} - \frac{1}{x-3}\right)$