Question

find the derivative of t^2/√t+1

Answer 1

quotient rule: low d high minus high d low all over low-low.

Answer 2

OR you can bring up the denominator since you should know:\[\frac{ 1 }{ a+1 }= (a+1)^{-1}\]

Answer 3

i used the quotient rule and i believe i am stuck because i stopped at \[\frac{ 2t-\frac{ t ^{2} }{ 2 } }{ (\sqrt{t+1})^{2} }\]

Answer 4

I think your derivative is wrong for the numerator. you forgot something next to the 2t.
\[(\sqrt{t}+1) \times 2t\]

Answer 5

and also, \[\frac{ 1 }{ \sqrt{t} } = t^{-\frac{ 1 }{ 2 }}\]

Answer 6

i simplified that once i written it out.

Answer 7

\[u=t^2\]
\[u"=2t\]
\[v=(t+1)^{\frac{ 1 }{ 2 }}\]
\[v"=\frac{ 1 }{ 2 }(t+1)^{-\frac{ 1 }{ 2 }}\]
\[=\frac{ 1 }{ 2\sqrt{t+1} }\]
\[y"=\frac{ (u"\times v)-(u \times v") }{ v^2 }\]
\[=\frac{ 2t \sqrt{t+1}-\frac{ t^2 }{ 2\sqrt{t+1} } }{ t+1 }\]
\[=\frac{ 4t(t+1)-t^2 }{ 2\sqrt{t+1} }\times \frac{ 1 }{ t+1 }\]
\[=\frac{ 4t^2+4t-t^2 }{ 2\sqrt{t+1} }\times \frac{ 1 }{ t+1 }\]

Answer 8

Continue that.

Answer 9

Make sure you read what I did.

Answer 10

And I used the quotient rule.