Question

how do you solve dy/dx for y=(x^2 + 3x)(x-2)(x^2 +1) by logarithmic differentiation?

Answer 1

so ln y = ln(x2+3x) + ln(x-2) + ln(x2 +1) right?

Answer 2

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

Answer 3

\[\ln(abc)=\ln a+\ln b+ \ln c\]

Answer 4

and we get 1/y (dy/dx) = [(1/ x2+3x)(2x+3)]+ [(1/ x-2)]+ [(1/ x2+1)(2x)]

Answer 5

im stuck here though...because do i just simplify? or do i have to do anything else?

Answer 6

dont just leave it as 1 on the nominator but derive the denominator at the top

Answer 7

\[\frac{dy}{dx}\ln(x^2+3x)=\dfrac{2x+3}{x^2+3x}\]

Answer 8

right so now we have \[[(2x+3)/(x ^{2} +3x)] +(1/x-2) + (2x/x ^{2} +1)\]

Answer 9

do i simplify more?

Answer 10

so multiply by y

Answer 11

wait why are we doing that?

Answer 12

since you have \[\frac{\frac{dy}{dx}}{y}\]

Answer 13

on the left

Answer 14

so multiply everything by y then?

Answer 15

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

Answer 16

ah ok

Answer 17

so now we have (x2 +3x)(x-2)(x2+1) + (x2+3x)(x2+1) + (2x)(x2+3x)(x-2)

Answer 18

sorry (2x+3)(x-2)(x2+1) + (x2+3x)(x2+1) + (2x)(x2+3x)(x-2)

Answer 19

should i leave it like that?

Answer 20

yes good work

Answer 21

i dont have to solve the binomials?

Answer 22

thanks for you help :)

Answer 23

your*

Answer 24

yw