Question

y=(1/x)^ln x

implicit differentation :(

Answer 1

is your expression as below:
\[\Large y = {\left( {\frac{1}{x}} \right)^{\ln x}}\]

Answer 2

\[y=(\frac{ 1 }{ x })^{\ln x}\]

Answer 3

yes :)

Answer 4

ok! Then we have to take the logarithm (in base e) of both sides, so we get:
\[\Large \ln y = \frac{{\ln x}}{x}\]

Answer 5

won't be a minute :)

Answer 6

ok!

Answer 7

sorry about that :)

Answer 8

no worries :)

Answer 9

I got \[-(\ln x)^{2}\]

Answer 10

I'm sorry, I don't think that it is the right answer @alekos

Answer 11

now we have to compute the first derivative of both sides of the last equation. In order to that the first derivative of the left side, is given applying the chain rule, so we get:
\[\Large \frac{{y'}}{y}\]

Answer 12

whereas in order to compute the first derivative of the right side, I apply the quotient rule, so I can write:
\[\Large \left( {\frac{{\ln x}}{x}} \right)' = \frac{{\frac{1}{x}x - \ln x}}{{{x^2}}}\]

Answer 13

is it ok?

Answer 14

yes got that :)

Answer 15

ok! so after the differentiation, we can write:
\[\Large \frac{{y'}}{y} = \frac{{\frac{1}{x}x - \ln x}}{{{x^2}}}\]

Answer 16

now, please note that we can simplify the right side as below:
\[\Large \frac{{\frac{1}{x}x - \ln x}}{{{x^2}}} = \frac{{1 - \ln x}}{{{x^2}}}\]

Answer 17

yes

Answer 18

so we can rewrite that expression above, as below:
\[\Large \frac{{y'}}{y} = \frac{{1 - \ln x}}{{{x^2}}}\]

Answer 19

now, in order to find the expresion of y', we have to multiply both sides by y, so what do you get?

Answer 20

expression*

Answer 21

Michele I'm a little confused how you got this :o\[\Large \ln y = \frac{{\ln x}}{x}\]Wouldn't log of each side give us this?\[\Large\rm \ln y=\ln x\cdot \ln\left(\frac{1}{x}\right)\]

Answer 22

\[y'=\frac{ y-\ln(xy) }{ x ^{2}y }\]

Answer 23

oops.. sorry you are right! @zepdrix I have made a big error
sorry @Kerriy

Answer 24

that is ok :)

Answer 25

Ahhh darn, I wish we had caught that earlier hehe :)

Answer 26

:) @zepdrix

Answer 27

we have to restart, so the right expression is:
\[\ln y = \ln x\ln \left( {\frac{1}{x}} \right) = \ln x \times \left( { - \ln x} \right) = - {\left( {\ln x} \right)^2}\]

Answer 28

so the first derivative of the right side is:
\[ \Large - 2\ln x\left( {\frac{1}{x}} \right)\]

Answer 29

then we can write:
\[\Large \frac{{y'}}{y} = - 2\ln x\left( {\frac{1}{x}} \right)\]

Answer 30

where does the -lnx come from ?

Answer 31

since, using the properties of logarithm we have:
\[\ln \left( {\frac{1}{x}} \right) = \ln 1 - \ln x = 0 - \ln x = - \ln x\]

Answer 32

ok thank you so much :)

Answer 33

then in order to find the value of y' we have to multiply bot sides by y, so we get:
\[\Large y' = - 2y\left( {\frac{1}{x}} \right)\ln x\]

Answer 34

You are a champion, I cannot begin to thank you enough :)
Can you tell me where I could go to learn more about logs and trig......or is it simply just practise?

Answer 35

you have to learn the basic properties, and then add much practise

Answer 36

Thank you again, I really appreciate your effort and time :)

Answer 37

thank you! :)
please in the last equation you ahve to replace y with its definition, namely:
\[\Large y = {\left( {\frac{1}{x}} \right)^{\ln x}}\]

Answer 38

so your answer is:
\[\Large y' = - 2{\left( {\frac{1}{x}} \right)^{\left( {\ln x + 1} \right)}}\ln x\]

Answer 39

Sorry again for my big error of before! @Kerriy

Answer 40

It is fine you have been wonderful :)

Answer 41

Thanks! :)

Answer 42

Welcome :)