| Thread Closed |
Differentiating this |
Share Thread | Thread Tools |
| Jun7-06, 03:19 AM | #1 |
|
|
Differentiating this
Hey. If i have:
[tex] y = 2x\sqrt {4 - 2x^3 } [/tex] To differentiate it, i used the product rule, but used the chain rule to differentiate the [itex]\sqrt {4 - 2x^3}[/itex] part. I got the answer right, but was just wondering, is there a quicker way of doing it? Or have i gone about it the right way? Thanks, Dan. |
| Jun7-06, 03:24 AM | #2 |
|
Recognitions:
|
You've done it the right way
|
| Jun7-06, 03:27 AM | #3 |
|
|
As far as I know there is no othere way (execept if you use a computer). And it really isn't that long either.
PS Somethimes it's easier if you differentiate using logoritems (especially if theres a lot of multiplication involved). |
| Jun7-06, 06:45 AM | #4 |
|
|
Differentiating this[tex]y= 2x(4- 2x^3)^\frac{1}{2}[/tex] at least makes the derivative a bit clearer. |
| Jun7-06, 07:29 AM | #5 |
|
|
|
| Jun7-06, 09:34 AM | #6 |
|
Recognitions:
|
Or you could do some algebra and throw the 2x into the square root before doing any calculus operations, which removes the need for the product rule.
[tex]y = 2x \sqrt{4-2x^3} = \sqrt{4x^2 (4-2x^3)} [/tex] |
| Jun7-06, 10:02 AM | #7 |
|
|
The above however doesn't work well, because you lose the sign.
Another potential way to simplify might be to take logarithm and differentiate that, but you have to be careful there. |
| Jun7-06, 06:44 PM | #8 |
|
|
Thanks for the replies everyone. |
| Thread Closed |
| Thread Tools | |
Similar Threads for: Differentiating this
|
||||
| Thread | Forum | Replies | ||
| Differentiating x^x | Precalculus Mathematics Homework | 10 | ||
| Differentiating | Calculus & Beyond Homework | 1 | ||
| Help Differentiating | Calculus & Beyond Homework | 4 | ||
| Differentiating Ln | Calculus & Beyond Homework | 7 | ||
| Differentiating | Calculus & Beyond Homework | 2 | ||