Logarithmic Differentiation

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Homework Statement



http://img253.imageshack.us/img253/5748/logdiffji4.jpg [Broken]

The Attempt at a Solution



I've tried both and i) and ii) however something just doesn't seem right. Can somebody tell me if my method is correct?

i) http://img253.imageshack.us/img253/8791/img0186anh3.jpg [Broken] ii)http://img329.imageshack.us/img329/627/img0187azr5.jpg [Broken]
 
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Answers and Replies

  • #2
Hurkyl
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You forgot to differentiate.
 
  • #3
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You forgot to differentiate.

I've got a total mindblock :yuck: , can you explain a little further?
 
  • #4
differentiate both right and left hand side...
 
  • #5
595
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differentiate both right and left hand side...

Oh yeh! I tried by using chain rule (Let sinx^cosx = u)
therefore i get :

(1/y)*(dy/dx) = (1/u)*[d/dx(sinx^cosx)]

How would I go about diff-ing sinx^cosx?
 
  • #6
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consider cosx as u and log(sinx) as v, then diff by the product rule (udv+vdu)
In diff. v, use the chain rule
 
  • #7
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consider cosx as u and log(sinx) as v, then diff by the product rule (udv+vdu)
In diff. v, use the chain rule

Ah! so much easier!
 
  • #8
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Ok, after that tip from f(x) I got what I beleive is the final answer. Can this be simplified any further?

http://img156.imageshack.us/img156/115/img0192us4.jpg [Broken]
 
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  • #9
D H
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Your work looks good up until very last step. Try that final step one more time.
 
  • #10
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Your work looks good up until very last step. Try that final step one more time.

silly me! :rofl:

2nd time lucky!
http://img441.imageshack.us/img441/8141/answerem9.jpg [Broken]
 
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  • #11
D H
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Much better.
 
  • #12
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I'm not too sure how to go about part (ii)
So far i've converted the square roots to ^1/2 and expanded out to get
x^(3/2)[(1-x^2)^(1/4)]

I then let that equal u (chain rule) however the product rule within u seems to get pretty messy/complicated.

Is this the correct method?
 
  • #13
cristo
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I'm not too sure how to go about part (ii)
So far i've converted the square roots to ^1/2 and expanded out to get
x^(3/2)[(1-x^2)^(1/4)]

I then let that equal u (chain rule) however the product rule within u seems to get pretty messy/complicated.

Is this the correct method?

If [itex]y=x^{3/2}[(1-x^2)^{1/4}],[/itex] then let [itex]u=x^{3/2} [/itex]and [itex]v=(1-x^2)^{1/4} [/itex] so that y'=uv'+u'v, remembering that to calculate v' you will need to use the chain rule.
 
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  • #14
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Ok, so I now know how to diff [itex]y=x^{3/2}[(1-x^2)^{1/4}],[/itex] but shouldn't I take log of both sides before diff-ing? In that case I need to figure out the derivative of ln[[itex]y=x^{3/2}[(1-x^2)^{1/4}],[/itex]]
 
  • #15
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Bump, can anybody help?
 
  • #16
HallsofIvy
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Ok, so I now know how to diff [itex]y=x^{3/2}[(1-x^2)^{1/4}],[/itex] but shouldn't I take log of both sides before diff-ing? In that case I need to figure out the derivative of ln[[itex]y=x^{3/2}[(1-x^2)^{1/4}],[/itex]]

You could but you don't have to. Typically "logarithmic differentiation is used when you have a function of x in the exponent.

Just use the product rule:
[itex]y'= (x^{3/2})'(1- x^2)^{1/4}+ x^{3/2}[(1-x^2)^{1/4}]'[/itex]
[itex]= (3/2)x^{1/2}(1- x^2)^{1/4}+ (x^{3/2})(1/4)(1-x^2)^{-3/4}[/itex]

If you take the logarithm of both sides, you get [itex]ln y= (3/2)ln(x)+ (1/4)ln(1+ x^2)[/itex]. Now [itex](1/y)y'= 3/(2x)+ 2x/(4(1+x^2))[/itex]. That righthand side is much simpler but you still have to multiply by y.
 
  • #17
VietDao29
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Another way to differentiate the second expression is to change it to a single n-root, like this:
[tex]\sqrt{x ^ 3 \sqrt{1 - x ^ 2}} = \sqrt{\sqrt{x ^ 6 (1 - x ^ 2)}} = \sqrt[4]{x ^ 6 - x ^ 8}[/tex]
Let u = x6 - x 8. Now, the whole expression becomes [tex]\sqrt[4]{u}[/tex], you can apply the chain rule and finish the problem. Can you go from here? :)
 
  • #18
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Another way to differentiate the second expression is to change it to a single n-root, like this:
[tex]\sqrt{x ^ 3 \sqrt{1 - x ^ 2}} = \sqrt{\sqrt{x ^ 6 (1 - x ^ 2)}} = \sqrt[4]{x ^ 6 - x ^ 8}[/tex]
Let u = x6 - x 8. Now, the whole expression becomes [tex]\sqrt[4]{u}[/tex], you can apply the chain rule and finish the problem. Can you go from here? :)

I think ill go with this method. So using chain rule I get dy/dx = (1/4)((x^6-x^8)^(-3/4))*(6x^5-8x^7)

Hope you can make that out. So where does the logarithmic part come into it? Do I now take log of both sides?

I'm quite lost:confused:
 
  • #19
VietDao29
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I think ill go with this method. So using chain rule I get dy/dx = (1/4)((x^6-x^8)^(-3/4))*(6x^5-8x^7)

Hope you can make that out. So where does the logarithmic part come into it? Do I now take log of both sides?

I'm quite lost:confused:
Nope, you can differentiate it normally as you've just done.

Or you can tuse logarithmic differentiation as HOI has suggested, i.e, it goes like this:

[tex]y = \sqrt{x ^ 3 \sqrt{1 - x ^ 2}}[/tex]

Taking log of both sides yields:

[tex]\ln y = \ln \left( \sqrt{x ^ 3 \sqrt{1 - x ^ 2}} \right) = \frac{1}{2} \ln \left( x ^ 3 \sqrt{1 - x ^ 2} \right) = \frac{1}{2} \left( \ln (x ^ 3) + \ln( \sqrt{1 - x ^ 2} ) \right)[/tex]

Now, you can differentiate both sides to find y'.

There are many ways to approach a differentiation problem, so just choose the one that you like best.
 
  • #20
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That's very helpful! Had to navigate through the diff of [tex]\frac{1}{2} \left( \ln (x ^ 3) + \ln( \sqrt{1 - x ^ 2} ) \right)[/tex]. Was a bit tricky but I think I got it...
http://img253.imageshack.us/img253/6788/asdfry0.jpg [Broken]

Can you please confirm?
 
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  • #21
VietDao29
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Nope, still incorrect. You forget to multiply it by y.

Multiply it by y, then simplify the expression you get, and it's all done. :)

Note that differentiating [tex]\ln( \sqrt{1 - x ^ 2} )[/tex] can be a little bit tricky, but if you pull out the 1/2, it should be easier. It goes like this:

[tex][\ln( \sqrt{1 - x ^ 2} ) ]' = \frac{1}{2} [\ln (1 - x ^ 2) ]' = \frac{1}{2} \times \frac{-2x}{1 - x ^ 2} = \frac{-x}{1 - x ^ 2}[/tex] :)
 
  • #22
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Nope, still incorrect. You forget to multiply it by y.

Multiply it by y, then simplify the expression you get, and it's all done. :)

Note that differentiating [tex]\ln( \sqrt{1 - x ^ 2} )[/tex] can be a little bit tricky, but if you pull out the 1/2, it should be easier. It goes like this:

[tex][\ln( \sqrt{1 - x ^ 2} ) ]' = \frac{1}{2} [\ln (1 - x ^ 2) ]' = \frac{1}{2} \times \frac{-2x}{1 - x ^ 2} = \frac{-x}{1 - x ^ 2}[/tex] :)

Yeh, that was just the diff without * y.

After making common denom of (6x^3)(4-4x^2) and * by y I get
http://img242.imageshack.us/img242/5305/finalansux3.jpg [Broken]
 
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  • #23
VietDao29
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Yeh, that was just the diff without * y.

After making common denom of (6x^3)(4-4x^2) and * by y I get
http://img242.imageshack.us/img242/5305/finalansux3.jpg [Broken]
[/URL]
You still haven't simplified it... =.="
HINT: You can factor out x2 in the numerator and denominator, then the two x2's will cancel each other out.

[tex]y \times \left( \frac{3x ^ 2}{6x ^ 3} - \frac{2x}{4 - 4 x ^ 2} \right) = \sqrt{x ^ 3 \sqrt{1 - x ^ 2}} \times \left( \frac{1}{2x} - \frac{x}{2 - 2 x ^ 2} \right)[/tex]

[tex]= \frac{1}{2}\sqrt{x ^ 3 \sqrt{1 - x ^ 2}} \times \left( \frac{1}{x} - \frac{x}{1 - x ^ 2} \right)[/tex]
And... you can stop here, and leave it in this form. :smile: It's okay, simplified enough. You shouldn't make common denominator, as it's unnecessary, and you may also make some minor mistakes during doing so.
 
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  • #24
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Finally got it. Many, many thanks!
 

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