Differentiating this

  • Thread starter danago
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  • #1
danago
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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.
 

Answers and Replies

  • #2
TD
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You've done it the right way :smile:
 
  • #3
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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).
 
  • #4
HallsofIvy
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danago said:
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.
 
  • #5
danago
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HallsofIvy said:
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.
 
  • #6
mezarashi
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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]
 
  • #7
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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.
 
  • #8
danago
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mezarashi said:
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 :rolleyes:

Thanks for the replies everyone.
 

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