Question

Differentiate: 3t^2-2sqrt(t)
(Calculus - Derivatives)

Answer 1

I'm stuck on this step

Answer 2

\[x=3t^2-2\sqrt{t}\]

Answer 3

do u have to do it with 1st principles?

Answer 4

Can you just take the derivative?

Answer 5

nope, we have to do it that way.

Answer 6

your works looks okay so far
divide that by \(\Delta t\) and take the limit

Answer 7

\[X(t) = 3t^2-2\sqrt{t}\] \[X'(t) = \lim\limits_{\Delta t \to 0} ~\dfrac{\Delta X}{\Delta t}\]

Answer 8

@ganeshie8 what about at this part.. what do i do with the ones with square roots?
\[\Delta(x)=6t \Delta t+3\Delta t^2 -2\sqrt{t+\Delta t}+2\sqrt{t}\]

Answer 9

lets see... \[\begin{align}X'(t) &= \lim\limits_{\Delta t \to 0} ~\dfrac{\Delta X}{\Delta t}\\~\\&=\lim\limits_{\Delta t \to 0} ~\dfrac{6t\Delta t + 3\Delta t^2 - 2\sqrt{t+\Delta t} + 2\sqrt{t}}{\Delta t}\\~\\\end{align}\]

Answer 10

you're stuck at that step, right ?

Answer 11

yeah, i get that you could factor out \[\Delta t\] so you can cancel it, but what about the one inside the square root symbol?

Answer 12

looks we need to handle it more carefully.. how about rationalizing ?

Answer 13

lets see... \[\begin{align}X'(t) &= \lim\limits_{\Delta t \to 0} ~\dfrac{\Delta X}{\Delta t}\\~\\&=\lim\limits_{\Delta t \to 0} ~\dfrac{6t\Delta t + 3\Delta t^2 - 2\sqrt{t+\Delta t} + 2\sqrt{t}}{\Delta t}\\~\\
&=\lim\limits_{\Delta t \to 0} ~6t+3\Delta t - \dfrac{ 2\sqrt{t+\Delta t} - 2\sqrt{t}}{\Delta t}\\~\\
\end{align}\]

Answer 14

try to rationalize the top part of that fraction

Answer 15

that just means we multiply it by a special kind of 1 : \(\large \dfrac{2\sqrt{t+\Delta t } + 2\sqrt{t}}{2\sqrt{t+\Delta t } + 2\sqrt{t}}\)

Answer 16

\[\begin{align}X'(t) &= \lim\limits_{\Delta t \to 0} ~\dfrac{\Delta X}{\Delta t}\\~\\&=\lim\limits_{\Delta t \to 0} ~\dfrac{6t\Delta t + 3\Delta t^2 - 2\sqrt{t+\Delta t} + 2\sqrt{t}}{\Delta t}\\~\\
&=\lim\limits_{\Delta t \to 0} ~6t+3\Delta t - \dfrac{ 2\sqrt{t+\Delta t} - 2\sqrt{t}}{\Delta t}\\~\\
&=\lim\limits_{\Delta t \to 0} ~6t+3\Delta t - \dfrac{ 2\sqrt{t+\Delta t} - 2\sqrt{t}}{\Delta t}\times \color{gray}{\dfrac{2\sqrt{t+\Delta t } + 2\sqrt{t}}{2\sqrt{t+\Delta t } + 2\sqrt{t}}} \\~\\
\end{align}\]

Answer 17

So you multiply it by it's conjugate, right? Now I get it. Thanks!

Answer 18

Yep! thats a very useful trick for getting rid of the evil \(\Delta t\) in the denominator

Answer 19

@ganeshie8 Thanks again.

Answer 20

yw