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September 18, 2014

September 18, 2014

Posted by **Anoymous** on Thursday, November 4, 2010 at 9:41pm.

F(x) = root5(3 + 5 x + x^3)?

- derivative -
**bobpursley**, Thursday, November 4, 2010 at 9:45pmf(x)=u^k

f'= ku^(k-1) du

so here, that does something like this..

I assume root5 means the 1/5 power.

f'= 1/5( )^(-4/5) * (5 + 3x^2)

- derivative -
**Anonymous**, Thursday, November 4, 2010 at 9:45pm5/2root(3+5x+x^3) * 5+3x^2=

(5(5+3x^2))/(2root(3+5x+x^2)

- derivative -
**Jack**, Thursday, November 4, 2010 at 9:48pmIs that the 5th root of (3+5x+x^3)? If so, you can rewrite it as (3+5x+x^3)^(1/5). You have to use the chain rule to find this derivative. First set (3+5x+x^3)=z. You know have z^(1/5). Take the derivative of this. (1/5)z^(1/5-1)=(1/5)z^(-4/5). Now substitute (3+5x+x^3) back in for z. (1/5)(3+5x+x^3)^(-4/5). Take the derivative of the inside, which is (3+5x+x^3). You get (0+5+3x^2)=(5+3x^2). Multiple this with (1/5)(3+5x+x^3)^(-4/5). You get (1/5)(5+3x^2)(3+5x+x^3)^(-4/5).

- derivative -
**Jack**, Thursday, November 4, 2010 at 9:50pmAnonymous is wrong, don't use that. bobpursley and I are correct.

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