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.
 
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> 'Whodunnit' of Irish potato famine solved
>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change
>> Curiosity Mars rover drills second rock target
Jun7-06, 03:24 AM   #2
TD
 
Recognitions:
Homework Helper Homework Help
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
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus

Differentiating this


Quote by danago
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.
I'm not sure what you would consider 'quicker' but writing this as
[tex]y= 2x(4- 2x^3)^\frac{1}{2}[/tex]
at least makes the derivative a bit clearer.
 
Jun7-06, 07:29 AM   #5
 
Quote by HallsofIvy
I'm not sure what you would consider 'quicker' but writing this as
[tex]y= 2x(4- 2x^3)^\frac{1}{2}[/tex]
at least makes the derivative a bit clearer.
Yea thats what i did. Then i said that [tex]f(x)=2x[/tex] and [tex]g(x)=(4- 2x^3)^\frac{1}{2}[/tex]. I then differentiated g(x) with the chain rule, then once i found that, i used the product rule to find the final derivative.
 
Jun7-06, 09:34 AM   #6
 
Recognitions:
Homework Helper Homework Help
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
 
Quote by mezarashi
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]
I never thought of doing it like that

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