Differentiating y = x^(1/3) (x - 4)^(2/3)

[Spoon-]

Yeah so I'm trying to differentiate y = x[sup:24xhm4g5]1/3[/sup:24xhm4g5] (x - 4)[sup:24xhm4g5]2/3[/sup:24xhm4g5]

I got x[sup:24xhm4g5]-2/3[/sup:24xhm4g5] (x-4)[sup:24xhm4g5]2/3[/sup:24xhm4g5] [1 + 2x(x-4)[sup:24xhm4g5]-1[/sup:24xhm4g5]] but I think I'm wrong.

 

[Dr. Flim-Flam]

Re: Differentiating y = x^(1/3) (x-4)^(2/3)

f(x) = x^(1/3)*(x-4)^(2/3) Use product rule.

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

= [x-4+2x]/[3x^(2/3)(x-4)^(1/3)] = (3x-4)/[3x^(2/3)(x-4)^(1/3)].

 

[Spoon-]

Re: Differentiating y = x^(1/3) (x-4)^(2/3)

Ok, is it possible to make that denominator into (3)(x[sup:3889j286]3[/sup:3889j286] - 4x[sup:3889j286]2[/sup:3889j286])[sup:3889j286]1/3[/sup:3889j286] ?

Also, what's the 2nd derivative of it?

I got: -32x / 9(x[sup:3889j286]3[/sup:3889j286] - 4x[sup:3889j286]2[/sup:3889j286])[sup:3889j286]-4/3[/sup:3889j286]

 

[soroban]

Re: Differentiating y = x^(1/3) (x-4)^(2/3)

Hello, Spoon!

\(\displaystyle \text{Ok, is it possible to make that denominator into: }\:3\left(x^3 - 4x^2\right)^{\frac{1}{3}}\) ? . . . . Yes!

Also, what's the 2nd derivative?

\(\displaystyle \text{I got: }\;\frac{-32x}{9\left(x^3 - 4x^2\right)^{\frac{4}{3}}}\) . . . . So did I!