Is My Logarithmic Differentiation Method Correct?

[t_n_p]

Homework Statement

http://img253.imageshack.us/img253/5748/logdiffji4.jpg

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 ii)http://img329.imageshack.us/img329/627/img0187azr5.jpg

 

[Hurkyl]

You forgot to differentiate.

 

[t_n_p]

You forgot to differentiate.

I've got a total mindblock , can you explain a little further?

 

[Ahmed Abdullah]

differentiate both right and left hand side...

 

[t_n_p]

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?

 

[f(x)]

consider cosx as u and log(sinx) as v, then diff by the product rule (udv+vdu)
In diff. v, use the chain rule

 

[t_n_p]

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!

 

[t_n_p]

Ok, after that tip from f(x) I got what I believe is the final answer. Can this be simplified any further?

http://img156.imageshack.us/img156/115/img0192us4.jpg

 

[D H]

Your work looks good up until very last step. Try that final step one more time.

 

[t_n_p]

Your work looks good up until very last step. Try that final step one more time.

silly me!

2nd time lucky!
http://img441.imageshack.us/img441/8141/answerem9.jpg

 

[D H]

Much better.

 

[t_n_p]

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?

 

[cristo]

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 y=x^{3/2}[(1-x^2)^{1/4}], then let u=x^{3/2}and v=(1-x^2)^{1/4} so that y'=uv'+u'v, remembering that to calculate v' you will need to use the chain rule.

 

[t_n_p]

Ok, so I now know how to diff y=x^{3/2}[(1-x^2)^{1/4}], but shouldn't I take log of both sides before diff-ing? In that case I need to figure out the derivative of ln[y=x^{3/2}[(1-x^2)^{1/4}],]

 

[t_n_p]

Bump, can anybody help?

 

[HallsofIvy]

Ok, so I now know how to diff y=x^{3/2}[(1-x^2)^{1/4}], but shouldn't I take log of both sides before diff-ing? In that case I need to figure out the derivative of ln[y=x^{3/2}[(1-x^2)^{1/4}],]

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:
y'= (x^{3/2})'(1- x^2)^{1/4}+ x^{3/2}[(1-x^2)^{1/4}]'
= (3/2)x^{1/2}(1- x^2)^{1/4}+ (x^{3/2})(1/4)(1-x^2)^{-3/4}

If you take the logarithm of both sides, you get ln y= (3/2)ln(x)+ (1/4)ln(1+ x^2). Now (1/y)y'= 3/(2x)+ 2x/(4(1+x^2)). That righthand side is much simpler but you still have to multiply by y.

 

[VietDao29]

Another way to differentiate the second expression is to change it to a single n-root, like this:
\sqrt{x ^ 3 \sqrt{1 - x ^ 2}} = \sqrt{\sqrt{x ^ 6 (1 - x ^ 2)}} = \sqrt[4]{x ^ 6 - x ^ 8}
Let u = x⁶ - x ⁸. Now, the whole expression becomes \sqrt[4]{u}, you can apply the chain rule and finish the problem. Can you go from here? :)

 

[t_n_p]

Another way to differentiate the second expression is to change it to a single n-root, like this:
\sqrt{x ^ 3 \sqrt{1 - x ^ 2}} = \sqrt{\sqrt{x ^ 6 (1 - x ^ 2)}} = \sqrt[4]{x ^ 6 - x ^ 8}
Let u = x⁶ - x ⁸. Now, the whole expression becomes \sqrt[4]{u}, 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

 

[VietDao29]

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

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:

y = \sqrt{x ^ 3 \sqrt{1 - x ^ 2}}

Taking log of both sides yields:

\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)

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.

 

[t_n_p]

That's very helpful! Had to navigate through the diff of \frac{1}{2} \left( \ln (x ^ 3) + \ln( \sqrt{1 - x ^ 2} ) \right). Was a bit tricky but I think I got it...
http://img253.imageshack.us/img253/6788/asdfry0.jpg

Can you please confirm?

 

[VietDao29]

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

[\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} :)

 

[t_n_p]

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

[\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} :)

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

 

[VietDao29]

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

[/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.

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)

= \frac{1}{2}\sqrt{x ^ 3 \sqrt{1 - x ^ 2}} \times \left( \frac{1}{x} - \frac{x}{1 - x ^ 2} \right)
And... you can stop here, and leave it in this form. 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.

 

[t_n_p]

Finally got it. Many, many thanks!