Question

please help with derivative below

Answer 1

\[\sqrt{2t}\]

Answer 2

i keep getting the anwer 1t^-(1/2)

Answer 3

because 2t^(1/2)

Answer 4

https://www.wolframalpha.com/input/?i=%E2%88%9A2t

Answer 5

there is a more simplifed answer that i am looking for but thank you

Answer 6

do this (2t)^(1/2) then take the derivative

Answer 7

power rule

Answer 8

yeah i did that and i got t^-1/2 but teacher marked it wrong

Answer 9

let y=sqrt (2t)
\[y=\sqrt{2}t ^{\frac{ 1 }{ 2 }},\]
\[\frac{ d }{ dx }\left( x \right)^n=nx ^{n-1}\]

Answer 10

\(\bf \sqrt{2t}\implies (2t)^{\frac{1}{2}}\qquad {\color{brown}{ chain \ rule}}
\\ \quad \\
\cfrac{d}{dx}\left[(2t)^{\frac{1}{2}}\right]\cdot \cfrac{d}{dx}\left[2t\right]\)

Answer 11

well, "t" rather for that matter =) \(\bf \cfrac{d}{dt}\left[(2t)^{\frac{1}{2}}\right]\cdot \cfrac{d}{dt}\left[2t\right]\)

Answer 12

@jdoe0001 i didnt know you had to use the chain rule. that makes sense. why the chain rule and not just 2t^1/2 and that give you 1t^-1/2

Answer 13

\[\frac{ dy }{ dt }=\sqrt{2}\frac{ 1 }{ 2}t ^{\frac{ -1 }{ 2}}=\frac{ 1 }{\sqrt{2t}}\]

Answer 14

\(\bf \sqrt{2t}\implies ({\color{brown}{ 2t}})^{\frac{1}{2}}\implies ({\color{brown}{ \blacksquare}})^{\frac{1}{2}}\) thus the chain rule

Answer 15

I mean chain rule not power rule

Answer 16

hmmm I guess one could also use the power rule \(\bf \sqrt{2t}\implies (2t)^{\frac{1}{2}}\implies 2^{\frac{1}{2}}\cdot t^{\frac{1}{2}}\implies \sqrt{2}\ t^{\frac{1}{2}}\)

Answer 17

thank all of you, i understand now