Question

using logarithmic differentiation find the derivative of the function:

Answer 1

\[\sqrt[6]{(x^2 +1)/(x^2 -1)}\]

Answer 2

You need a y=

Answer 3

yeah y = that

Answer 4

ok introduce ln to y and your function.

Answer 5

i have \[\ln y = 1/6 \ln (x^2+1)/(x^2-1)\]

Answer 6

Ok now remember,
\[\ln(a/b) = ln a - ln b\]

Answer 7

so it's \[\ln y = 1/6 \ln(x^2+1) - \ln(x^2-1) \] ?

Answer 8

The 1/6 should be multiplied to both functions but we can factor it to avoid any mistakes.

Answer 9

\[lny= 1/6( \ln(x^2+1) - \ln (x^2-1))\]

Answer 10

Now can you go on?

Answer 11

no note quite.....

Answer 12

ok,\[ d/dx lnx= 1/g(x) \times g'(x)\] right?

Answer 13

\[d/dx (lnx)=1/g(x)×g′(x)\]

Answer 14

yes d/dx [ln g(x)] = g'(x)/g(x)

Answer 15

ok so differentiate each term using that rule first and remember
the derivative of y is y' since we are looking for dy/dx.
Then let me know what you get and we can move on to the next step.

Answer 16

so would it be \[\ln y =2x/[(x^2+1)/(x^2-1)] d/dx[(x^2+1)/(x^2-1)] \] all times 1/6 ?

Answer 17

I'm not sure why you have a d/dx there. But this should be simple.
Let me do half the question for you.
derivative of\[ lny = 1/y \times dy/dx\]
Derivative of\[ \ln (x^2+1) = 1/(x^2+1) \times 2x\]
Derivative of\[ \ln (x^2-1) = 1/(x^2-1) \times 2x\]
Go on from here and let me know what you have as dy/dx

Answer 18

\[d/dx[\ln(x^2+1)-\ln(x^2-1)]1/6 = 1/6[(2x/2x+1)-(2x/x^2-1)]\]

Answer 19

Hold on, It's \[x^2+1\] not \[2x+1\].
Also, the question is not yet finished because
\[1/y \times dy/dx= (2x/x^2+1)- (2x/x^2-1)\]
So we are meant to solve for dy/dx. What can you do to solve for dy/dx?

Answer 20

i can use chain rule to make \[dy/dx = dy/du *du/dx = 1/u du/dx\]

Answer 21

No you do not have to, since we are looking for dy/dx.
we just multiply everything by y to get \[dy/dx=1/6y(2x/x2+1)−(2x/x2−1)\]
But still we are not done.

Answer 22

i'm sorry i'm still completely clueless

Answer 23

ok, when they ask you to differentiate using logs, they want you to find dy/dx

Answer 24

So what you do is IMPLICITLY differentiate everything.

Answer 25

remember implicit differentiation?

Answer 26

i'm trying to. my book isn't being of much help

Answer 27

You see y' is the same as dy/dx

Answer 28

In the question I just typed, to find y' or dy/dx.
We would have to differentiate all terms.

Answer 29

Find y' if
\[2=2x^3y^2+xy\] <-- this was my question.
Let's start with the left hand side, derivative of 2 is 0
Then we move on to the right hand side

Answer 30

ok so that's dydx 2= 2x^3 * y^2 +xy
is 0 = 6x^2*2y+1

Answer 31

we have to functions that need to be derived using product rule.

Answer 32

two*

Answer 33

That's incorrect.
Remember product rule=
\[f'g + fg'\]

Answer 34

\[0 = 6x^2y^2 + 2x^3*2y + y+x\]

Answer 35

Ok now this is almost correct.
Just remember, every time you are looking for y' or dy/dx when you differentiate y you get y' or dy/dx
So the derivative of y^2= 2yy'
and derivative of y= y'

Answer 36

Whenever you differentiate y add a y' or dy/dx

Answer 37

ok so it's \[0=6x^2y^2 + 2x^3*2yy' +y'+x\]

Answer 38

and we are to isolate y'

Answer 39

exactly and then solve for y'.

Answer 40

But then it would be \[0=6x^2y^2+2x^3(2yy)+y+xy'\]

Answer 41

sorry that's 2yy'

Answer 42

\[(-6x^2y^2 - x )/ (2x^3*2y*2) = y'\]

Answer 43

ok then with that infon then it's different from that

Answer 44

I don''t think you isolated y' correctly. 6x^2y^2 and y would go to the left hand side of the equation so that you are left with terms that contain y' only. Then just factor out y'
Try it again.

Answer 45

i still get \[(-6x^2y^2 -y)/(2x^3+x+2y+2) = y'\]

Answer 46

You are almost write. Are you isolating the equation I wrote down or yours?
Because you were missing and xy' and y. Compare the two equations.
But anyway, seems like you got the point.
Now we apply the same knowledge to the log function.

Answer 47

the one you wrote....yeah i think i get it mostly now. how do i go about applying it to the log eqn?

Answer 48

Ok so you got everything until the part where we simplified the log right?
Because before you do anything to do with log differentiation, always simplify the function.

Answer 49

yes i have \[1/y*dy/dx=(2x/x^2 +1)−(2x/x^2 −1) \]

Answer 50

Great, don't forget the 1/6

Answer 51

So now just solve for dy/dx as we did in the previous example.

Answer 52

ok so i muliply the function by y to give [dy/dx=1/6y(2x/x2+1)−(2x/x2−1)\] right?

Answer 53

yes but then you would have to plug in your original function because remember.
\[y= \sqrt[6]{(x2+1)/(x2−1)}\]

Answer 54

so it would be \[1/6(\sqrt[6]{(x^2+1/x^2-1}) (2x/x^2+1)-(2x/x^2-1)\] ?

Answer 55

It's 1/6y multiplied to the whole function so multiply \[1/6(\sqrt[6]{(x2+1/x2−1)} )\] to all of this ---->\[(2x/x^2+1)−(2x/x^2−1)\]
to get \[1/6(\sqrt[6]{(x2+1/x2−1)} (2x/x^2+1)−(2x/x^2−1))\]

Answer 56

that looks like what i put, is it different?

Answer 57

Because \[dy/dx= 1/6y'(2x/x^2+1)−1/6y'(2x/x^2−1)\]
So we just factor out the 1/6 y'

Answer 58

\[1/6 \sqrt[6]{1/6((x2+1/x2−1)} ((2x/x2+1)−(2x/x2−1))\]
This is your answer, notice I have double parenthesis to show that my 1/6y' is multiplied to both functions in parenthesis?

Answer 59

so it is just a bit different from yours, but you get the point I see.

Answer 60

I meant that is the correct answer^^ or rather the answer you need.

Answer 61

so that one is the final answer or one that still needs to be simplified?

Answer 62

That is your final answer, simplifying would just be tedious.

Answer 63

\[ y' =( 24x/x^2 - 1 ) \sqrt[6]{(x^2+1)/(x^2-1)}\] would that translate into

Answer 64

i think it might giving the factoring of the 6y' and the two 2x's

Answer 65

The two 2x's have a different base so you would have to find a common denominator in order to add them.

Answer 66

Also, you were missing 1/6 in that answer.

Answer 67

i meant x^4 - 1 in that first part

Answer 68

Your common denominator is write, but adding the 2x's would give you -4x wouldn't it? divided by \[6(x^4-1)\] because of the 1/6.
So you would end up with -\[2x/3x^4-3\]

Answer 69

right*

Answer 70

\[-2x/3x^4-3\]

Answer 71

you are right! thank you so much! you help al lot

Answer 72

No problem. Just do lots of practice and you will be used to this in no time.

Answer 73

:) you deserve all your medals

Answer 74

haha ,you too, for your patience and interest in learning.