posted by Anoymous .
How do you find the derivative of
F(x) = root5(3 + 5 x + x^3)?
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)
5/2root(3+5x+x^3) * 5+3x^2=
Is 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).
Anonymous is wrong, don't use that. bobpursley and I are correct.