Question

Use Logarithmic Differentiation to solve:
y= [(x+1)^3 * (x-2)^3]/3(x^3 -5)^1/2

If there is a way to show a photo of the equation I can upload that. I've tried this multiple times and can seem to get the answer given :/
ANS > ( y' = {ln|x+1|+ln|x-2|-(x^2)/2 * ln|x^3 -5|]*y )

Answer 1

[y=\frac{ (x+1)^{3} \times (x-2)^{3}}{ 3\sqrt{x ^{3}-5} }\]

Answer 2

\[y=\frac{ (x+1)^{3} \times (x-2)^{3}}{ 3\sqrt{x ^{3}-5} }\]

Answer 3

How did you post it as a equation?

Answer 4

and thank you!

Answer 5

Use the 'Equation'

Answer 6

I meant the button

Answer 7

then after inserting it it comes out in the worded form

Answer 8

Do you get
\[\ln(y)=\ln[(x+1)^3(x-2)^3]-\ln(3(3x^3-5)^{1/2}\]

Answer 9

yes that's what I get

Answer 10

but after that I differentiate each term and get a completely different answer

Answer 11

Do you get
\[\ln(y)=3\ln(x^2-x-2)-\ln(3)-\frac{1}{2}\ln(x^3-5)\]

Answer 12

how do you get that line? normally wouldn't it be

\[\ln (y)= \ln (x+1)^{3} + \ln (x-2)^{3} - \ln (3\times \sqrt{x ^{3}-5}\]

Answer 13

\[\ln(y)=\ln[(x+1)^3(x-2)^3]-\ln(3(3x^3-5)^{1/2}) \\ \\ \ln(y)=\ln[(x+1)^3(x-2)^3]-[\ln(3)+\frac{1}{2}\ln(x^3-5)]\]

Answer 14

then how does the first term become

\[3\times \ln (x ^{2}-x-2)\]

Answer 15

\[\ln[(x+1)^3(x-2)^3] \\ \\ \ln[(x+1)(x-2)]^3 \\ \\ 3\ln[(x+1)(x-2)]\]

Answer 16

Then differentiate it

Answer 17

\[\frac{1}{y}\frac{dy}{dx}=\frac{3}{x^2-x-2}(2x-1)-\frac{1}{2}(\frac{1}{x^3-5})(3x^2)\]

Answer 18

so after differentiating should there be a?

\[\frac{ x ^{2} }{ 2 }\]

Answer 19

this is the answer I have on the answer sheet

Answer 20

This is what you get after differentiating
\[\frac{1}{y}\frac{dy}{dx}=\frac{3}{x^2-x-2}(2x-1)-\frac{1}{2}(\frac{1}{x^3-5})(3x^2)\]

Answer 21

Then
\[\frac{1}{y}\frac{dy}{dx}=\frac{3}{x^2-x-2}(2x-1)-\frac{1}{2}(\frac{1}{x^3-5})(3x^2)\]
\[\frac{dy}{dx}=y[\frac{3}{x^2-x-2}(2x-1)-\frac{1}{2}(\frac{1}{x^3-5})(3x^2)]\]
Substitute y in it

Answer 22

They want it in logarithmic form I see

Answer 23

Wait I gotta go

Answer 24

Good luck

Answer 25

@saifoo.khan

Answer 26

Where he stopped?

Answer 27

Well they want it in logarithmic form and I'm not sure how to get it?

Answer 28

Oh. Hold on a sec.

Answer 29

I'm sorry. I can't figure it out. :(

Answer 30

ah that's alright, thanks for trying
I shall have to wait until Sam comes back later