# Logarithmic Differentiation

1. Apr 21, 2007

### t_n_p

1. The problem statement, all variables and given/known data

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

3. 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]

Last edited by a moderator: May 2, 2017
2. Apr 21, 2007

### Hurkyl

Staff Emeritus
You forgot to differentiate.

3. Apr 21, 2007

### t_n_p

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

4. Apr 21, 2007

### Ahmed Abdullah

differentiate both right and left hand side...

5. Apr 21, 2007

### t_n_p

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. Apr 21, 2007

### 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

7. Apr 21, 2007

### t_n_p

Ah! so much easier!

8. Apr 21, 2007

### t_n_p

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]

Last edited by a moderator: May 2, 2017
9. Apr 21, 2007

### D H

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

10. Apr 21, 2007

### t_n_p

silly me! :rofl:

2nd time lucky!

Last edited by a moderator: May 2, 2017
11. Apr 21, 2007

### D H

Staff Emeritus
Much better.

12. Apr 21, 2007

### 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?

13. Apr 21, 2007

### cristo

Staff Emeritus
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.

Last edited: Apr 21, 2007
14. Apr 22, 2007

### 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[$y=x^{3/2}[(1-x^2)^{1/4}],$]

15. Apr 23, 2007

### t_n_p

Bump, can anybody help?

16. Apr 23, 2007

### HallsofIvy

Staff Emeritus
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.

17. Apr 23, 2007

### 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? :)

18. Apr 24, 2007

### t_n_p

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

19. Apr 24, 2007

### VietDao29

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.

20. Apr 25, 2007

### 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 [Broken]