Find the Derivative of F(t)= Suare root t / 2t =3

[math1325]

Find the Derivative of F(t)= Suare root t / 2t =3

My answer so far...

(2t+3)(square root t)-(square root t)*(2t+3)
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divided by (2t+3)^2


then (2t+3)1/2squaret-squaret(2t)
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divided by (2t+3)^2


then (2t+3)-4tsquare
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devided by 2square root (2t+3)62


then 2t-1tsquare
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divided by 2 square root (2t+3)square


then t-tsquare
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divided by square root (2t+3)square


at last i get t
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divided by square root (2t+3)square

 

[soroban]

Hello, math13251

That's not the Quotient Rule used on this planet . . .

Find the derivative of: \(\displaystyle \L f(t)\;=\;\frac{\sqrt{t}}{2t\,+\,3} \;=\;\frac{t^{\frac{1}{2}}}{2t\,+\,3}\)

\(\displaystyle \text{We have: }\L\,f'(t)\;=\;\frac{(2t\,+\,3)\cdot\frac{1}{2}\cdot t^{-\frac{1}{2}}\,-\,t^{\frac{1}{2}}\cdot 2}{(2t\,+\,3)^2}\)

\(\displaystyle \text{Multiply top and bottom by }2t^{\frac{1}{2}}:\L\;\;f'(t)\;=\;\frac{2t\,+\,3\,-\,4t}{2t^{\frac{1}{2}}(2t\,+\,3)^2}\)

Therefore: \(\displaystyle \L\,f'(t)\;=\;\frac{3\,-\,2t}{2\sqrt{t}(2t\,+\,3)^2}\)