Question

Differentiate the function.

Answer 1

\[v=(\sqrt{x}+(1\div \sqrt[3]{x}))^{2}\]

Answer 2

\[v=\left( \sqrt{x}+\frac{ 1 }{ \sqrt[3]{x} } \right)^{2}\]
\[v \prime=2\left( \sqrt{x}+\frac{ 1 }{ \sqrt[3]{x} } \right)\frac{ d }{ dx }\left( x ^{\frac{ 1 }{ 2 }} +x ^{\frac{ -1 }{ 3 }}\right)\]
can you complete it?

Answer 3

\[v=(x ^{1/2}+x ^{-1/2})^{2}\]
\[v=(x ^{1/2}+x ^{-1/2})(x ^{1/2}+x ^{-1/2})\]
\[v=(x+x ^{1/6}+x ^{1/6}+x ^{-2/3}\]
am i on the right track??

Answer 4

no,it is \[x ^{\frac{ -1 }{ 3 }}\]

Answer 5

In your way
\[v=\left( x ^{\frac{ 1 }{ 2 }}+x ^{\frac{ -1 }{ 3 }} \right)^2=x+x ^{\frac{ -2 }{ 3 }}+2 x ^{\frac{ 1 }{ 2 }}x ^{\frac{ -1 }{ 3 }}\]
\[or~v=x+x ^{\frac{ -2 }{ 3 }}+2x ^{\frac{ 3-2 }{ 6 }}=x+x ^{\frac{ -2 }{ 3 }}+2 x ^{\frac{ 1 }{ 6 }}\]
v'=?