Question

differentiate y = ln sqrt(e^3x (x^2 - 1))

Answer 1

\[\ln \sqrt{e ^{3x}(x ^{2}-1)}\]

Answer 2

use logarrithm properties.
\(\ln \sqrt{x}=(1/2)ln x \\ ln(ab)=ln a+ ln b \\ ln e =1\)

Answer 3

so first step would be \[\frac{ \ln (e ^{3x}) + \ln(x^{2}-1) }{ 2 }\]

Answer 4

your first step is correct.

Answer 5

what is the lne = 1?

Do i then use the \[\ln f(x) = \frac{ f \prime (x) }{ f(x) }\] on the top two functions?
What happens to the 2 at the bottom though?

Answer 6

e is the base of natural logarithm, so logarithm of e = 1 ---> ln e =1
ln (e^(3x)) = 3x ln e = 3x.
got this ?

Answer 7

yes i see that now.

Answer 8

keep bottom 2 as it is....till the end.

Answer 9

so \[\frac{ 3x + \ln(x ^{2}-1) }{ 2 }\] is next step?

Answer 10

yup.
can u differentiate that?
or need further help?

Answer 11

ill try now if i cant ill post a msg.

Answer 12

sure :)

Answer 13

is it \[\frac{ 3 + \frac{ 2x }{ x ^{2}-1 } }{ 2 }\]

Answer 14

thats correct :)
simplify it....

Answer 15

\[\frac{ 3+ \frac{ 2x }{ (x+1)(x-1) } }{ 2 }\]

Answer 16

i meant
3(x^2-1)+2x in the numerator.........

Answer 17

\(\huge\frac{ 3 + \frac{ 2x }{ x ^{2}-1 } }{ 2 }=\frac{3(x^2-1)+2x}{2(x^2-1)}\)
simplify the numerator.

Answer 18

pardon?

Answer 19

oh wait i see

Answer 20

do I need to then simplify that further?

Answer 21

yes,
3x^2+2x-3
in numerator.

Answer 22

that would be it.....
(3x^2+2x-3 ) / 2(x^2-1)

Answer 23

thank you very much for your help. I got very confused as I have a program that suplies the answers and the answer to this question came to
\[\frac{ \frac{ 3(x ^{2}-1)e ^{3x} }{ 2 } + xe ^{3x} }{ (x ^{2}-1)e ^{3x} }\]

Answer 24

does that make any sense?

Answer 25

i wonder how there still is e^(3x) term....

Answer 26

thats mainly what confused me with this. Its Microsoft mathematics. Its worked for other sums but this one it went around the bend a little.

Answer 27

is it explicitly mentioned to use product rule ??

Answer 28

nope.

Answer 29

i think they didn't simplify first, directly differentiated and used chain rule....

Answer 30

is it wrong to do it that way?

Answer 31

not wrong,but complicated....we found an equivalent answer in much simpler way.....

Answer 32

yes thank you very much for that. Btw is it possible for you to explain the chain rule briefly?

Answer 33

example :
\(\large \frac{d}{dx}ln (sin x)=\frac{1}{sin x}\quad\frac{d}{dx}(sin x)=cos x/sinx=cot x.\)

Answer 34

how does sin become cos/sin? surely f'(x) of sin = cos?

Answer 35

sin did not become cos / sin
derivative of ln(sin x) became cos / sin
derivative of ln x = 1/ x
so derivative of ln sin x = ?

Answer 36

derivative of ln x = x'/x

Answer 37

yup, thats chain rule.

Answer 38

isnt chain rule f'(g(x)).g'(x)?

Answer 39

same thing.
g(x)=x,f(x)=ln x
u just wrote the defination of derivative of f(g(x))

Answer 40

ahhhhhh i see now. thank you. what if you have more than two functions though?

Answer 41

f(g(h(x)))= f'(g(h(x))). g'(h(x)). h'(x)

Answer 42

now that makes more sense than my lecturer lol.

Answer 43

glad to hear that :)

Answer 44

thank you very much.