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Use of logrithmic differentiation
- Thread starter paul davo
- Start date
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
Full Member
- Joined
- Oct 25, 2012
- Messages
- 598
You may find the a uTube Video useful.
Hello, paul davo!
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 . . . \(\displaystyle \displaystyle y' \;=\;\frac{(x+1)(x-2)^3}{x-3}\left(\frac{1}{x+1} + \frac{3}{x-2} - \frac{1}{x-3}\right) \)
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 . . . \(\displaystyle \displaystyle y' \;=\;\frac{(x+1)(x-2)^3}{x-3}\left(\frac{1}{x+1} + \frac{3}{x-2} - \frac{1}{x-3}\right) \)