Question

Product rule for x^(4/5) (x-4)^2

Answer 1

product rule:\[(uv)'=u'v+uv'\]

u = x^(4/5) v = (x-4)^2

u' = (4/5)x^(-1/5) "using power rule"

v' = 2x-8 "using chain rule"


\[(uv)' = (\frac{ 4 }{ 5 }x^{-\frac{1}{5}})(x-4)^2+(2x-8)(x^{\frac{4}{5}})\]
\[= (\frac{ 4 }{ 5\sqrt[5]{x} })(x^2-8x+16)+(\sqrt[5]{x^4})(2x-8)\]
\[\frac{ 4x^2-32x+64 }{ 5\sqrt[5]{x} }+2x \sqrt[5]{x^4}-8\sqrt[5]{x^4}\]
\[\frac{ 4x^2-32x+16 }{ \sqrt[5]{3125x^5} }+\sqrt[5]{32x^9}-\sqrt[5]{32768x^4}\]

Answer 2

oh that 16 at the bottom should be 64

Answer 3

you can choose whatever version of the derivative you like

Answer 4

i suggest the first one cuz its simpler